题干：

You are given an array aa of nn integers, where nn is odd. You can make the following operation with it:

• Choose one of the elements of the array (for example aiai) and increase it by 11(that is, replace it with ai+1ai+1).

You want to make the median of the array the largest possible using at most kkoperations.

The median of the odd-sized array is the middle element after the array is sorted in non-decreasing order. For example, the median of the array [1,5,2,3,5][1,5,2,3,5] is 33.

Input

The first line contains two integers nn and kk (1≤n≤2⋅1051≤n≤2⋅105, nn is odd, 1≤k≤1091≤k≤109) — the number of elements in the array and the largest number of operations you can make.

The second line contains nn integers a1,a2,…,ana1,a2,…,an (1≤ai≤1091≤ai≤109).

Output

Print a single integer — the maximum possible median after the operations.

Examples

Input

3 2
1 3 5

Output

5

Input

5 5
1 2 1 1 1

Output

3

Input

7 7
4 1 2 4 3 4 4

Output

5

Note

In the first example, you can increase the second element twice. Than array will be [1,5,5][1,5,5] and it's median is 55.

In the second example, it is optimal to increase the second number and than increase third and fifth. This way the answer is 33.

In the third example, you can make four operations: increase first, fourth, sixth, seventh element. This way the array will be [5,1,2,5,3,5,5][5,1,2,5,3,5,5] and the median will be 55.

题目大意：

  给你n个数，让你可以最多执行K次操作，每次操作使得一个数+1，问你操作完之后的序列的中位数最大是多大。

解题报告：

  水题，排序后发现只跟后一般的数有关。考虑每次操作，操作前半个区间的数字没有意义。操作后面的数字的话也没有意义，因为看的是中位数，所以只有中间这个数字变大，答案才能变大。（也就是说使得答案变大的方式只有使得中间这个数字变大）然后模拟这个过程就行了。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
ll a[MAX];
map<ll,ll> mp;
set<ll> ss;
int main()
{int n,k;cin>>n>>k;for(int i = 1; i<=n; i++) cin>>a[i];sort(a+1,a+n+1);for(int i = (n+1)/2; i<=n; i++) ss.insert(a[i]),mp[a[i]]++;ll num = mp[a[(n+1)/2]];//当前这个大点集的点的个数 ll ans = a[(n+1)/2];while(k>0) {auto it = ss.upper_bound(ans);if(it == ss.end()) break;ll ci = (*it-ans)*num;if(ci <= k) {k -= ci;ans = *it;num += mp[*it];}else {ans += (k / (num));k = 0;	}}if(k > 0) ans += k/(num);cout << ans << endl;return 0 ;
}
//16:16-16:26

总结：刚开始WA了一发，是因为else中k=0和ans+=(k/sum)这两句写反了，，，真服了自己了。