C. Maximum Median - 技术栈

time limit per test

2 seconds

memory limit per test

256 megabytes

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

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

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

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] is 3.

Input

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

The second line contains n integers a1,a2,...,an (1≤ai≤109).

Output

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

Examples

Input

Copy

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3 2
1 3 5

Output

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Input

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5 5
1 2 1 1 1

Output

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Input

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7 7
4 1 2 4 3 4 4

Output

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Note

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

In the second example, it is optimal to increase the second number and than increase third and fifth. This way the answer is 3.

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] and the median will be 5.

解题说明：此题是一道数学题，首先对数列进行从小到大排序，为了让中位数变大,让中位数变成和后面的数一样大,于是让该中位数在加完k后和后面的X个数字相同即可。

cpp 复制代码

#include<iostream>
#include<algorithm>
using namespace std;
int n, k, m, i;
int main()
{
	cin >> n >> k;
	int a[200004];
	m = n / 2;
	for (i = 0; i < n; i++)
	{
		cin >> a[i];
	}
	sort(a, a + n);
	for (i = a[m]; k > 0; i++) 
	{
		while (i == a[m + 1])
		{
			m++;
		}
		k = k - (m - n / 2 + 1);
	}
	if (k == 0)
	{
		cout << i<<endl;
	}
	else
	{
		cout << i - 1<<endl;
	}
	return 0;
}