Codeforces Round #496 (Div. 3) E2. Median on Segments (General Case Edition) 中位数为m的区间的个数

E2. Median on Segments (General Case Edition)

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given an integer sequence $a_1,a_2,\dots ,a_n$

.

Find the number of pairs of indices $(l,r)$

( $1\le l\le r\le n$) such that the value of median of $a_l,a_{l+1},\dots ,a_r$ is exactly the given number $m$

.

The median of a sequence is the value of an element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if $a=[4,2,7,5]$

then its median is $4$ since after sorting the sequence, it will look like $[2,4,5,7]$ and the left of two middle elements is equal to $4$. The median of $[7,1,2,9,6]$ equals $6$ since after sorting, the value $6$

will be in the middle of the sequence.

Write a program to find the number of pairs of indices $(l,r)$

( $1\le l\le r\le n$) such that the value of median of $a_l,a_{l+1},\dots ,a_r$ is exactly the given number $m$

.

Input

The first line contains integers $n$

and $m$ ( $1\le n,m\le 2\cdot {10}^5$

) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence $a_1,a_2,\dots ,a_n$

( $1\le a_i\le 2\cdot {10}^5$

).

Output

Print the required number.

Examples

Input

Copy

5 4
1 4 5 60 4

Output

Copy

8

Input

Copy

3 1
1 1 1

Output

Copy

6

Input

Copy

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

Output

Copy

97

Note

In the first example, the suitable pairs of indices are: $(1,3)$

, $(1,4)$, $(1,5)$, $(2,2)$, $(2,3)$, $(2,5)$, $(4,5)$ and $(5,5)$.

/**
题意 给定长度为n的序列的子序列中 中位数为m的个数
思路: solved(n,m):表示中位数小于等于m的区间数 solved(n,m-1) :表示中位数小于等于m-1的区间的个数;
那么相减之后就是ans;
对于solve(n,m)的求法 可利用+1 -1 来进行运算 
枚举右端点 计算 大于0 的数量  累加 树状数组维护即可;


*/

#include<bits/stdc++.h>
#define ll unsigned long long
using namespace std;

const int maxn=2e5+7;
int a[maxn],b[2*maxn];

struct node{
	int c[maxn*2],n;
	void init(int _n){
		n=_n;
		for(int i=0;i<=n;i++) c[i]=0;
	}
	void add(int st){
		for(int i=st;i<=n;i+=i&-i) c[i]++;
	}
	int sum(int st){
		int res=0;
		for(int i=st;i>0;i-=i&-i) res+=c[i];
		return res;
	}
}tmp;

ll solved(int n,int m){
	for(int i=1;i<=n;i++) b[i]=(a[i]<=m?1:-1);
	for(int i=1;i<=n;i++) b[i]+=b[i-1];
	tmp.init(n*2+1);
	tmp.add(n+1);
	ll res=0;
	for(int i=1;i<=n;i++){
		res+=tmp.sum(n+1+b[i]);
		tmp.add(b[i]+n+1);
	}
	return res;
}

int main(){
	int n,m;scanf("%d %d",&n,&m);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	printf("%I64d\n",solved(n,m)-solved(n,m-1));
	return 0;
}