Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3 6
7
5
7

Sample Output 1

5

All the non-empty contiguous subsequences of a are listed below:

{a_1} = {7}

{a_1, a_2} = {7, 5}

{a_1, a_2, a_3} = {7, 5, 7}

{a_2} = {5}

{a_2, a_3} = {5, 7}

{a_3} = {7}

Their means are 7, 6, 19/3, 5, 6 and 7, respectively, and five among them are 6 or greater. Note that {a_1} and {a_3} are indistinguishable by the values of their elements, but we count them individually.

Sample Input 2

1 2
1

Sample Output 2

0

Sample Input 3

7 26
10
20
30
40
30
20
10

Sample Output 3

13

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)

using namespace std;
typedef long long int ll;
typedef pair <ll, int> P;

const int MAX_N=2*100000;
ll bit[MAX_N+10], a[MAX_N+1],s[MAX_N+1];
P p[MAX_N+1];

ll sum(ll i){
    ll s=0;
    while(i>0){
        s+=bit[i];
        i-=i&-i;
    }
    return s;
}

void add(ll i, int x){
    while(i<=MAX_N){
        bit[i]+=x;
        i+=i&-i;
    }
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    ll N,K;
    cin>>N>>K;
    rep(i,0,N){
        cin>>a[i];
        a[i]-=K;
    }
    s[0]=0;
    p[0]=P(0,0);
    rep(i,0,N){
        s[i+1]=s[i]+a[i];
        p[i+1]=P(s[i+1],i+1);
    }
    sort(p,p+N+1);
    int temp=1;
    a[p[0].second]=temp;
    rep(i,1,N+1){
        if (p[i].first!=p[i-1].first){
            temp++;
        }
        a[p[i].second]=temp;
    }
    ll ans=0;
    rep(i,0,N+1){
        ans+=sum(a[i]);
        add(a[i],1);
    }
    cout << ans<<"\n";
}