#include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 10;
int n, m;
int sum[mx << 1];
int a[mx];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {
  while (x <= 2 * n + 1) {
    sum[x] += v;
    x += lowbit(x);
  }
}
int query(int x) {
  int ans = 0;
  while (x) {
    ans += sum[x];
    x -= lowbit(x);
  }
  return ans;
}
int main() {
  while (scanf("%d %d", &n, &m) != EOF) {
    long long x, y;
    x = y = 0;
    int cnt = n + 1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    memset(sum, 0, sizeof(sum));
    add(cnt, 1);
    for (int i = 1; i <= n; i++) {
      if (a[i] < m)
        cnt--;
      else
        cnt++;
      x += query(cnt - 1);
      add(cnt, 1);
    }
    memset(sum, 0, sizeof(sum));
    cnt = n + 1;
    add(cnt, 1);
    for (int i = 1; i <= n; i++) {
      if (a[i] <= m)
        cnt--;
      else
        cnt++;
      y += query(cnt - 1);
      add(cnt, 1);
    }
    printf("%lld\n", x - y);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

You are given an integer sequence a_1, a_2, ..., a_n.

Find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

The median of a sequence is the value of an element which is in the middle of the sequence after sorting it in non-decreasing order. If the length of the sequence is even, the left of two middle elements is used.

For example, if a=[4, 2, 7, 5] then its median is 4 since after sorting the sequence, it will look like [2, 4, 5, 7] and the left of two middle elements is equal to 4. The median of [7, 1, 2, 9, 6] equals 6 since after sorting, the value 6 will be in the middle of the sequence.

Write a program to find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

Input

The first line contains integers n and m (1 ≤ n,m ≤ 2⋅10^5) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2⋅10^5).

Output

Print the required number.

Examples

Input

5 4
1 4 5 60 4


Output

8


Input

3 1
1 1 1


Output

6


Input

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3


Output

97

Note

In the first example, the suitable pairs of indices are: (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 5), (4, 5) and (5, 5).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mx = 2e5 + 10;
int n, m;
int sum[mx << 1];
int a[mx];
int lowbit(int x) { return x & (-x); }
void add(int x, int v) {
  while (x <= 2 * n + 1) {
    sum[x] += v;
    x += lowbit(x);
  }
}
int query(int x) {
  int ans = 0;
  while (x) {
    ans += sum[x];
    x -= lowbit(x);
  }
  return ans;
}
int main() {
  while (scanf("%d %d", &n, &m) != EOF) {
    long long x, y;
    x = y = 0;
    int cnt = n + 1;
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    memset(sum, 0, sizeof(sum));
    add(cnt, 1);
    for (int i = 1; i <= n; i++) {
      if (a[i] < m)
        cnt--;
      else
        cnt++;
      x += query(cnt - 1);
      add(cnt, 1);
    }
    memset(sum, 0, sizeof(sum));
    cnt = n + 1;
    add(cnt, 1);
    for (int i = 1; i <= n; i++) {
      if (a[i] <= m)
        cnt--;
      else
        cnt++;
      y += query(cnt - 1);
      add(cnt, 1);
    }
    printf("%lld\n", x - y);
  }
  return 0;
}

```