CodeForce--Benches

A. Benches

 

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

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

Nobody leaves the taken seat during the whole process.

Input

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

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

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

Output

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

Examples

input

Copy

4
6
1
1
1
1

output

Copy

3 7

input

Copy

1
10
5

output

Copy

15 15

input

Copy

3
6
1
6
5

output

Copy

6 12

input

Copy

3
7
1
6
5

output

Copy

7 13

Note

In the first example, each of four benches is occupied by a single person. The minimum kk is 33. 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 kk is 77. That requires all six new people to occupy the same bench.

The second example has its minimum kk equal to 1515 and maximum kk equal to 1515, as there is just a single bench in the park and all 1010people will occupy it.

题目大意：公园里有n个座位，一开始每个座位都有多少人（人数给出），公园里将进来m人，问这些人往座位上坐，那么座位上人数最多的那个能有多少人，最多人数最小化能有多少人？　　

　　自己还是太菜了啊，看了一下大佬的题解，发现思维真的太强了

　　最多能有多少肯定好计算，把数组排序加上将要进来的人数就是，也就是代码中的kmax，问题是如何求得最小化的最大值呢？

　　大佬还是先排序，排好序了，让数组下标从0开始，依次与数组最大值比较（这个比较好理解，因为要是不超过的话肯定数组最大值就是最大啊），

　　然后，如果第i个元素和最大值相等，那么这个元素自增，并且m要自减，下面再来一个循环，只要这个元素比最大值小，那么这个元素就进行自增，m进行自减

　　知道m为0或者这个元素不小于最大值了，如果这轮i==n了但是m还没有为0，那么此时就要重新将i赋值为0

　　结束这层循环以后，要在外层循环中再次对数组进行排序，这个目的是为了更新最大值,最后输出就ok了

#include<iostream>
#include<algorithm>
using namespace std;
int arr[];
int n,m;
int kmax;
void solve()
{
    int i=;
    sort(arr,arr+n);
    kmax=arr[n-]+m;
    while(m!=)
    {
    if(arr[i]==arr[n-]&&m!=)
    {
        arr[i]++;
        m--;
    }
    while(arr[i]<arr[n-]&&m!=)
    {
        arr[i]++;
        m--;
    }
    i++;
    if(i==n)
    i=;
    }
    i=;
    sort(arr,arr+n);//更新最大值
}
int main()
{
    while(cin>>n>>m)
    {
        for(int i=;i<n;i++)
        cin>>arr[i];
        solve();
        cout<<arr[n-]<<" "<<kmax<<endl;
    }
    return ;
}