CF1042A Benches

版权声明：大佬您能赏脸，蒟蒻倍感荣幸，还请联系我让我好好膜拜。 https://blog.csdn.net/ShadyPi/article/details/82751508

原题链接：http://codeforces.com/contest/1042/problem/A

Benches

There are $n$ benches in the Berland Central park. It is known that $a_i$ people are currently sitting on the $i$-th bench. Another $m$ people are coming to the park and each of them is going to have a seat on some bench out of $n$ available.

Let $k$ be the maximum number of people sitting on one bench after additional $m$ people came to the park. Calculate the minimum possible $k$ and the maximum possible $k$.

Nobody leaves the taken seat during the whole process.

Input

The first line contains a single integer $n(1≤n≤100)$ — the number of benches in the park.

The second line contains a single integer $m(1≤m≤10000)$ — the number of people additionally coming to the park.

Each of the next $n$ lines contains a single integer $a_i(1≤a_i≤100)$ — the initial number of people on the $i$-th bench.

Output

Print the minimum possible $k$ and the maximum possible $k$, where $k$ is the maximum number of people sitting on one bench after additional $m$ people came to the park.

Examples

input

4
6
1
1
1
1

output

3 7

input

1
10
5

output

15 15

input

3
6
1
6
5

output

6 12

input

3
7
1
6
5

output

7 13

Note

In the first example, each of four benches is occupied by a single person. The minimum $k$ is $3$. For example, it is possible to achieve if two newcomers occupy the first bench, one occupies the second bench, one occupies the third bench, and two remaining — the fourth bench. The maximum $k$ is $7$. That requires all six new people to occupy the same bench.

The second example has its minimum $k$ equal to $15$ and maximum $k$ equal to $15$, as there is just a single bench in the park and all $10$people will occupy it.

题解

我都不知道我为什么要排序 （可能是身处排序强校的缘故吧）。

代码

#include<bits/stdc++.h>
using namespace std;
int a[105],maxn,n,m;
void in(){scanf("%d%d",&n,&m);for(int i=1;i<=n;++i)scanf("%d",&a[i]);}
void ac()
{
	sort(a+1,a+1+n);maxn=a[n]+m;
	for(int i=1;i<=n;++i)m-=a[n]-a[i];
	if(m<0)printf("%d",a[n]);
	else printf("%d",a[n]+m/n+(bool)(m%n));
	printf(" %d",maxn);
}
int main(){in();ac();}