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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

template<typename T> struct BinaryIndexedTree {
private : 
    vector<T> data;
public :
    BinaryIndexedTree(int n) : data(n+1, 0) { }

    T get(int i) {
        T ret = 0;
        for (; i >= 0; i = (i&(i+1))-1) ret += data[i];
        return ret;
    }

    void add (int i, T x) {
        for (; i < data.size(); i = (i|(i+1))) data[i] += x;
    }

};

int main() {
    int n;
    long long k;
    cin >> n >> k;
    vector<long long> a(n), sum(n+1, 0);
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        a[i] -= k;
        sum[i+1] = sum[i] + a[i];
    }
    
    vector<long long> v = sum;
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    auto idx = [&] (int x) -> int {
        return lower_bound(v.begin(), v.end(), x) - v.begin();
    };
    
    long long ans = 0;
    BinaryIndexedTree<long long> bit(n+5);
    for (const auto& e : sum) {
        int id = idx(e);
        ans += bit.get(id);
        bit.add(id, 1);
    }
    cout << ans << endl;
    return 0;
}