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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.TreeMap;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] sa = br.readLine().split(" ");
		int n = Integer.parseInt(sa[0]);
		int k = Integer.parseInt(sa[1]);
		long[] a = new long[n];
		for (int i = 0; i < n; i++) {
			a[i] = Integer.parseInt(br.readLine()) - k;
		}
		br.close();

		TreeMap<Long, Integer> map = new TreeMap<>();
		map.put(a[0], 0);
		for (int i = 1; i < n; i++) {
			a[i] += a[i - 1];
			map.put(a[i], 0);
		}

		Long[] arr = map.keySet().toArray(new Long[0]);
		int cnt = 0;
		for (Long i : arr) {
			cnt++;
			map.put(i, cnt);
		}
		int[] b = new int[n];
		for (int i = 0; i < n; i++) {
			b[i] = map.get(a[i]);
		}

		long ans = 0;
		BIT bit = new BIT(map.size());
		for (int i = 0; i < n; i++) {
			ans += bit.sum(b[i]);
			if (a[i] >= 0) {
				ans++;
			}
			bit.add(b[i], 1);
		}
		System.out.println(ans);
	}

	static class BIT {
		int n;
		long[] arr;

		public BIT(int n) {
			this.n = n;
			arr = new long[n + 1];
		}

		void add(int idx, long val) {
			for (int i = idx; i <= n; i += i & -i) {
				arr[i] += val;
			}
		}

		long sum(int idx) {
			long sum = 0;
			for (int i = idx; i > 0; i -= i & -i) {
				sum += arr[i];
			}
			return sum;
		}
	}
}