Codeforces Round #490 (Div. 3)--Mishka and Contest，Reversing Encryption，Alphabetic Removals

A. Mishka and Contest

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Mishka started participating in a programming contest. There are $n$ problems in the contest. Mishka's problem-solving skill is equal to $k$.

Mishka arranges all problems from the contest into a list. Because of his weird principles, Mishka only solves problems from one of the ends of the list. Every time, he chooses which end (left or right) he will solve the next problem from. Thus, each problem Mishka solves is either the leftmost or the rightmost problem in the list.

Mishka cannot solve a problem with difficulty greater than $k$. When Mishka solves the problem, it disappears from the list, so the length of the list decreases by $1$. Mishka stops when he is unable to solve any problem from any end of the list.

How many problems can Mishka solve?

Input

The first line of input contains two integers $n$ and $k$ ($1\le n,k\le 100$) — the number of problems in the contest and Mishka's problem-solving skill.

The second line of input contains $n$ integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le 100$), where $a_i$ is the difficulty of the $i$-th problem. The problems are given in order from the leftmost to the rightmost in the list.

Output

Print one integer — the maximum number of problems Mishka can solve.

Examples

input

Copy

8 4
4 2 3 1 5 1 6 4

output

Copy

5

input

Copy

5 2
3 1 2 1 3

output

Copy

0

input

Copy

5 100
12 34 55 43 21

output

Copy

5

Note

In the first example, Mishka can solve problems in the following order: 

$[4,2,3,1,5,1,6,4]\to [2,3,1,5,1,6,4]\to [2,3,1,5,1,6]\to [3,1,5,1,6]\to [1,5,1,6]\to [5,1,6]$

, so the number of solved problems will be equal to  5

In the second example, Mishka can't solve any problem because the difficulties of problems from both ends are greater than $k$.

In the third example, Mishka's solving skill is so amazing that he can solve all the problems.

思路：只需要两边分别判断即可，然后循环。AC代码：

#include<stdio.h>
int n,k;
int a[110];
int main()
{
	scanf("%d%d",&n,&k);
	for(int i=0;i<n;i++) scanf("%d",&a[i]);
	int i=0,j=n-1;
	int sum=0;
	while(i<=j)
	{
		if(a[i]<=k) sum++,i++;
		else if(a[j]<=k) sum++,j--;
		else if(a[i]>k && a[j]>k) break;
	}
	printf("%d\n",sum);
	return 0;
}

B. Reversing Encryption

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A string $s$ of length $n$ can be encrypted by the following algorithm:

• iterate over all divisors of $n$ in decreasing order (i.e. from $n$ to $1$),
• for each divisor $d$, reverse the substring $s[1\dots d]$ (i.e. the substring which starts at position $1$ and ends at position $d$).

For example, the above algorithm applied to the string $s$="codeforces" leads to the following changes: "codeforces" $\to$ "secrofedoc" $\to$ "orcesfedoc" $\to$ "rocesfedoc" $\to$ "rocesfedoc" (obviously, the last reverse operation doesn't change the string because $d=1$).

You are given the encrypted string $t$. Your task is to decrypt this string, i.e., to find a string $s$ such that the above algorithm results in string $t$. It can be proven that this string $s$ always exists and is unique.

Input

The first line of input consists of a single integer $n$ ($1\le n\le 100$) — the length of the string $t$. The second line of input consists of the string $t$. The length of $t$ is $n$, and it consists only of lowercase Latin letters.

Output

Print a string $s$ such that the above algorithm results in $t$.

Examples

input

Copy

10
rocesfedoc

output

Copy

codeforces

input

Copy

16
plmaetwoxesisiht

output

Copy

thisisexampletwo

input

Copy

1
z

output

Copy

z

Note

The first example is described in the problem statement.

思路：按照题目的意思模拟即可，每次判断当前数是否可以整除n。AC代码：

#include<stdio.h>
#include<string.h>
int n;
char s[110];
int a[10];
void f(int i,int j)
{
	while(i<j)
	{
		char c=s[i];
		s[i]=s[j];
		s[j]=c;
		i++;
		j--;
	}
}
int main()
{
	scanf("%d",&n);
	scanf("%s",s+1);
	for(int i=1;i<=n;i++)
	{
		if(n%i) continue;
		f(1,i);
	}
	printf("%s\n",s+1);
	return 0;
}

C. Alphabetic Removals

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a string $s$ consisting of $n$ lowercase Latin letters. Polycarp wants to remove exactly $k$ characters ($k\le n$) from the string $s$. Polycarp uses the following algorithm $k$ times:

• if there is at least one letter 'a', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
• if there is at least one letter 'b', remove the leftmost occurrence and stop the algorithm, otherwise go to next item;
• ...
• remove the leftmost occurrence of the letter 'z' and stop the algorithm.

This algorithm removes a single letter from the string. Polycarp performs this algorithm exactly $k$ times, thus removing exactly $k$ characters.

Help Polycarp find the resulting string.

Input

The first line of input contains two integers $n$ and $k$ ($1\le k\le n\le 4\cdot {10}^5$) — the length of the string and the number of letters Polycarp will remove.

The second line contains the string $s$ consisting of $n$ lowercase Latin letters.

Output

Print the string that will be obtained from $s$ after Polycarp removes exactly $k$ letters using the above algorithm $k$ times.

If the resulting string is empty, print nothing. It is allowed to print nothing or an empty line (line break).

Examples

input

Copy

15 3
cccaabababaccbc

output

Copy

cccbbabaccbc

input

Copy

15 9
cccaabababaccbc

output

Copy

cccccc

input

Copy

1 1
u

output

Copy

思路：先统计每个字母的个数，然后判断k在哪个字母的位置即可完成删减。AC代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
int n,k;
string s;
int a[27];
int main()
{
	cin>>n>>k;
	cin>>s;
	memset(a,0,sizeof(a));
	for(int i=0;i<s.size();i++)
	{
		int t=s[i]-96;
		a[t]++;
	}
	int x=-1;
	for(int i=1;i<27;i++)
	{
		if(k<=a[i])
		{
			x=i;
			break;
		}
		k-=a[i];
	}
	for(int i=0;i<s.size();i++)
	{
		int t=s[i]-96;
		if(t<x) continue;
		if(t==x && k)
		{
			k--;
			continue;
		}
		cout<<s[i];
	}
	cout<<endl;
	return 0;
}