999A - Mishka and Contest

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.

题意：一个人解决n个问题，这个问题的值比k小，每次只能解决最左边的或者最右边的问题，解决了就消失了。问这个人能解决多少个问题。

题解：模拟  可以先从左向右开始找小于等k的，然后删除掉，否则跳出循环，然后从右边向左找小于等于k的，删掉，大于k跳出循环。还可以不用删除，直接找小于的。

#include<bits/stdc++.h>
using namespace std;
int n,k,a[110],b[110],ans;
int main()
{
    cin>>n>>k;
    for(int i=0; i<n; i++)
        cin>>a[i];
    for(int i=0; i<n; i++)
        if(a[i]<=k)
            ans++,b[i]=1;
            else break;
    for(int i=n-1; i>=0; i--)
        if(a[i]<=k&&!b[i])
            ans++;
            else break;
    cout<<ans;
    return 0;
}

n,k=map(int,input().split())
a=list(map(int,input().split()))
ans=0
while a and a[0]<=k:
    a.pop(0);ans+=1
a.reverse()
while a and a[0]<=k:
    a.pop(0);ans+=1
print (ans)