You are given an integer sequence of length N, a = {a_1, a_2, …, a_N}, and an integer K.

a has N(N+1)/2 non-empty contiguous subsequences, {a_l, a_{l+1}, …, a_r} (1 ≤ l ≤ r ≤ N). Among them, how many have an arithmetic mean that is greater than or equal to K?

Constraints

* All input values are integers.
* 1 ≤ N ≤ 2 \times 10^5
* 1 ≤ K ≤ 10^9
* 1 ≤ a_i ≤ 10^9

Input

Input is given from Standard Input in the following format:


N K
a_1
a_2
:
a_N


Output

Print the number of the non-empty contiguous subsequences with an arithmetic mean that is greater than or equal to K.

Examples

Input

3 6
7
5
7


Output

5


Input

1 2
1


Output

0


Input

7 26
10
20
30
40
30
20
10


Output

13
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 200005
ll sum[N],ls[N];
ll lsree[N];
int n,k,nn,c;
void add(int x)
{
	while(x<=nn)//nn为上限，要大于离散化后的最大的数字
	{
		lsree[x]++;
		x+=x&-x;   //向上更新
	}
}
ll query(int x)
{
	ll s=0;
	while(x)
	{
		s+=lsree[x];
		x-=x&-x;   //向下求和
	}
	return s;
}
int main()
{
	scanf("%d %d",&n,&k);
	nn=n+5;
	for(int i=1;i<=n;i++)
	{
		int a;
		scanf("%d",&a);
		ls[i]=sum[i]=sum[i-1]+a-k;   //预处理出sum[i]数组
	}
	sort(ls,ls+1+n);   //排序，排n+1个数字的序（包括了sum[0]）
	c=unique(ls,ls+1+n)-ls;   //去重
	for(int i=0;i<=n;i++)
	sum[i]=lower_bound(ls,ls+1+c,sum[i])-ls+1;  //离散化 
	
	ll s=0;
	for(int i=0;i<=n;i++)
	{
		s+=query(sum[i]);  //查询有多少个小于sum[i]的数字
		add(sum[i]);   //更新 
	}
	printf("%lld\n",s);
	return 0;
}