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 <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>

using namespace std;

typedef long long ll;

ll n, k;
ll a[200005];
ll b[200005];
ll d[200005];
ll ans;
vector<ll> ve;

void f(ll p)
{
    while(p <= n)
    {
        d[p]++;
        p += (p & (-p));
    }
}

ll g(ll p)
{
    ll re = 0;
    while(p > 0)
    {
        re += d[p];
        p -= (p & (-p));
    }
    return re;
}

int main()
{
    scanf("%lld%lld", &n, &k);
    b[0] = 1;
    for(ll i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
        a[i] -= k;
        b[i] = b[i - 1] + a[i];
        ve.push_back(b[i]);
    }
    sort(ve.begin(), ve.end());
    ve.erase(unique(ve.begin(), ve.end()), ve.end());
    for(ll i = 0; i <= n; i++)
    {
        b[i] = lower_bound(ve.begin(), ve.end(), b[i]) - ve.begin() + 1;
        ans += g(b[i]);
        f(b[i]);
    }
    printf("%lld\n", ans);
    return 0;
}