1037_B. Reach Median
You are given an array a of n integers and an integer s. It is guaranteed that n is odd.

In one operation you can either increase or decrease any single element by one. Calculate the minimum number of operations required to make the median of the array being equal to s.

The median of the array with odd length is the value of the element which is located on the middle position after the array is sorted. For example, the median of the array 6, 5, 8 is equal to 6, since if we sort this array we will get 5, 6, 8, and 6 is located on the middle position.

Input

The first line contains two integers n and s (1≤ n≤ 2⋅ 10^5-1, 1≤ s≤ 10^9) — the length of the array and the required value of median.

The second line contains n integers a_1, a_2, …, a_n (1≤ a_i ≤ 10^9) — the elements of the array a.

It is guaranteed that n is odd.

Output

In a single line output the minimum number of operations to make the median being equal to s.

Examples

Input

3 8
6 5 8


Output

2

Input

7 20
21 15 12 11 20 19 12


Output

6

Note

In the first sample, 6 can be increased twice. The array will transform to 8, 5, 8, which becomes 5, 8, 8 after sorting, hence the median is equal to 8.

In the second sample, 19 can be increased once and 15 can be increased five times. The array will become equal to 21, 20, 12, 11, 20, 20, 12. If we sort this array we get 11, 12, 12, 20, 20, 20, 21, this way the median is 20.
```python

# -*- coding: utf-8 -*-
# @Date    : 2018-09-03 08:46:01
# @Author  : raj lath (oorja.halt@gmail.com)
# @Link    : http://codeforces.com/contest/1037/problem/B
# @Version : 1.0.0

import os
from sys import stdin

max_val=int(10e12)
min_val=int(-10e12)

def read_int()     : return int(stdin.readline())
def read_ints()    : return [int(x) for x in stdin.readline().split()]
def read_str()     : return input()
def read_strs()    : return [x for x in stdin.readline().split()]
def read_str_list(): return [x for x in stdin.readline().split().split()]

nb_elemets, value_needed = read_ints()
elements = sorted(read_ints())
mid = nb_elemets//2
ans = abs(elements[mid] - value_needed)
ans += sum( max(0, (a - value_needed)) for a in elements[:mid] )
ans += sum( max(0, (value_needed - a)) for a in elements[mid+1:] )
print(ans)




```