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>
using namespace std;

const int N = (int) 2e5 + 5;
const long long INF = (long long) 1e18;

int n, k;
long long s[N], fw[N];

void compress() {
	vector <long long> zip; zip.push_back(-INF);
	for (int i = 0; i <= n; i++) {
		zip.push_back(s[i] - 1LL * i * k);
	}
	sort(zip.begin(), zip.end());
	zip.resize(distance(zip.begin(), unique(zip.begin(), zip.end())));
	for (int i = 0; i <= n; i++) {
		s[i] = lower_bound(zip.begin(), zip.end(), s[i] - 1LL * i * k) - zip.begin();
	}
}

void upd (int i) {
	for (; i <= n + 1; i += i & -i) fw[i]++;
}

long long get(int i) {
	long long res = 0;
	for (; i >= 1; i -= i & -i) res += fw[i];
	return res;
}

int main() {
	//freopen(".inp", "r", stdin); freopen(".out", "w", stdout);
	scanf("%d %d", &n, &k);
	for (int i = 1; i <= n; i++) {
		int v; scanf("%d", &v);
		s[i] = s[i - 1] + v;
	}
	compress();
	long long ans = 0;
	upd(s[0]);
	for (int i = 1; i <= n; i++) {
		ans += get(s[i]);
		upd(s[i]);
	}
	printf("%lld\n", ans);
	return 0;
}