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 <iostream>
#include <ccomplex>
#include <vector>
#include <cassert>
#include <utility>
#include <algorithm>
#include <string>
#include <deque>
#include <queue>
#include <functional>
#include <cmath>
#include <iomanip>
#include <map>
#include <numeric>
#include <list>
#include <assert.h>
#include <math.h>
#include <valarray>
#include <stdio.h>
#include <algorithm>
#include <set>
#include <complex>
#include <list>
#include <time.h>
#include <stack>
#include <locale>
#include <clocale>
#include <ctype.h>
#include <wchar.h>
#include <chrono>

using namespace std;
typedef long long int LL;
typedef pair<long long int, long long int> pii;
typedef pair<double, double> pdd;

#define SORT(c) sort((c).begin(),(c).end())
#define BACKSORT(c) sort((c).begin(),(c).end(),std::greater<LL>())
#define FOR(i,a,b) for(LL i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)

int xcomp(vector<LL>& a, vector<LL>& b) {
	return a[0] < b[0];
}

#define XSORT(c) sort((c).begin(),(c).end(),xcomp)

#define ALL(x) (x).begin(),(x).end()
LL mod = 1000000007;

vector<LL> v(1, 0);

LL insert(LL n) {
	LL ans = 0;
	bool flag = false;
	for (LL j = 100000; j >= 10; j /= 10) {
		for (LL i = 0; i <= j; i += j / 10) {
			if (ans + i >= v.size()) {
				ans += i - j / 10;
				break;
			}
			if (n<v[ans + i]) {
				ans += i;
				break;
			}
		}
	}
	v.insert(v.begin() + ans, n);
	return ans;
}

int main() {
	v.push_back(9000000000000000000);
	LL N, K;
	cin >> N >> K;
	vector<LL> a(N);
	LL sum = 0;
	LL ruiseki = 0;
	REP(i, N) {
		cin >> a[i];
		a[i] -= K;
		ruiseki += a[i];
		sum += insert(ruiseki);
	}
	cout << sum << endl;
}