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 <vector>
#include <algorithm>
using namespace std;

typedef long long ll;
typedef pair<ll, int> pi;

int N, K;
int V;
vector<pi> B;
vector<int> C;
vector<ll> bit;

void init(int n) {
	V = 1;
	while (V < n)
		V <<= 1;
}

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

void add(int i, int x) {
	while (i <= V) {
		bit[i] += x;
		i += i & (-i);
	}
}

int main(void) {
	cin.tie(0);
	ios::sync_with_stdio(false);

	cin >> N >> K;
	B.resize(N + 1, make_pair(0LL, 0));
	for (int i = 1; i <= N; ++i) {
		int a;
		cin >> a;
		B[i].first = B[i - 1].first + a - K;
		B[i].second = i;
	}

	C.resize(N + 1);
	sort(B.begin(), B.end());
	int idx = -1;
	for (int i = 0; i <= N; ++i) {
		if (idx < 0)
			idx = i + 1;
		else if (B[i].first != B[i - 1].first)
			idx = i + 1;
		C[B[i].second] = idx;
	}

	init(N + 1);
	bit.resize(V + 1, 0);
	ll ans = 0;
	for (int i = 0; i <= N; ++i) {
		ans += sum(C[i]);
		add(C[i], 1);
	}

	cout << ans << '\n';
	return 0;
}