CF1042A Benches

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

Benches

There are nnn benches in the Berland Central park. It is known that aia_iai people are currently sitting on the iii-th bench. Another mmm people are coming to the park and each of them is going to have a seat on some bench out of nnn available.

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

Nobody leaves the taken seat during the whole process.

Input

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

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

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

Output

Print the minimum possible kkk and the maximum possible kkk, where kkk is the maximum number of people sitting on one bench after additional mmm 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 kkk is 333. 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 kkk is 777. That requires all six new people to occupy the same bench.

The second example has its minimum kkk equal to 151515 and maximum kkk equal to 151515, as there is just a single bench in the park and all 101010people 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();}

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CF1042A Benches

Benches

Input

Output

Examples

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output

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output

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Note

题解

代码

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