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 ll long long
#define fr first
#define sc second
using namespace std;

const int mnx = 1e6 + 9;
const int mod = 1e9 + 7;

ll n, k, a[mnx], t[mnx], h;
pair <ll, ll> z[mnx];

ll get(int x){
    ll res = 0;
    for(; x > 0; x -= (x & -x)){
        res += t[x];
    }
    return res;
}

void upd(int x){
    for(; x <= n; x += (x & -x)){
        t[x]++;
    }
}

void solve(){
    cin >> n >> k;

    for(int i = 1; i <= n; i++){
        cin >> a[i];
        a[i] -= k;
    }

    ll sum = 0;

    for(int i = 1; i <= n; i++){
        sum += a[i];
        z[i] = {sum, i};
    }

    sort (z + 1, z + n + 1);

    ll ans = 0;

    for(int i = 1; i <= n; i++){
        upd(z[i].sc);
        ans += get(z[i].sc);
        if(z[i].fr < 0) ans--;
    }

    cout << ans << '\n';
}


int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    ll tt = 1;

//    scanf("%d", &tt);

    while(tt--){
        solve();
    }

    return 0;
}