#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int inf = 1e9;
int a[maxn], num1[maxn], num2[maxn];
map<int, int> mp1, mp2;
int main() {
  int n, m, i, j, k, x, y, z, pos, min1, min2, max1, max2;
  long long ans, cnt;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == m) pos = i;
    }
    num1[0] = num2[0] = 0;
    mp1.clear();
    mp2.clear();
    min1 = min2 = inf;
    max1 = max2 = -inf;
    for (i = pos - 1; i > 0; i--) {
      if (a[i] < m)
        x = -1;
      else
        x = 1;
      num1[pos - i] = num1[pos - i - 1] + x;
      mp1[num1[pos - i]]++;
      min1 = min(min1, num1[pos - i]);
      max1 = max(max1, num1[pos - i]);
    }
    for (i = pos + 1; i <= n; i++) {
      if (a[i] < m)
        x = -1;
      else
        x = 1;
      num2[i - pos] = num2[i - pos - 1] + x;
      mp2[num2[i - pos]]++;
      min2 = min(min2, num2[i - pos]);
      max2 = max(max2, num2[i - pos]);
    }
    ans = 1 + mp1[0] + mp2[0] + mp1[1] + mp2[1];
    cnt = 0;
    for (i = min2; i <= max2; i++) {
      cnt += (long long)((long long)mp1[-i] + mp1[-i + 1]) * mp2[i];
    }
    for (i = min1; i <= max1; i++) {
      cnt += (long long)((long long)mp2[-i] + mp2[-i + 1]) * mp1[i];
    }
    printf("%lld\n", ans + cnt / 2);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

You are given a permutation p_1, p_2, ..., p_n. A permutation of length n is a sequence such that each integer between 1 and n occurs exactly once in the sequence.

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

The median of a sequence is the value of the 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 the median of p_l, p_{l+1}, ..., p_r is exactly the given number m.

Input

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

The second line contains a permutation p_1, p_2, ..., p_n (1 ≤ p_i ≤ n). Each integer between 1 and n occurs in p exactly once.

Output

Print the required number.

Examples

Input

5 4
2 4 5 3 1


Output

4


Input

5 5
1 2 3 4 5


Output

1


Input

15 8
1 15 2 14 3 13 4 8 12 5 11 6 10 7 9


Output

48

Note

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 2e5 + 10;
const int inf = 1e9;
int a[maxn], num1[maxn], num2[maxn];
map<int, int> mp1, mp2;
int main() {
  int n, m, i, j, k, x, y, z, pos, min1, min2, max1, max2;
  long long ans, cnt;
  while (scanf("%d%d", &n, &m) != EOF) {
    for (i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      if (a[i] == m) pos = i;
    }
    num1[0] = num2[0] = 0;
    mp1.clear();
    mp2.clear();
    min1 = min2 = inf;
    max1 = max2 = -inf;
    for (i = pos - 1; i > 0; i--) {
      if (a[i] < m)
        x = -1;
      else
        x = 1;
      num1[pos - i] = num1[pos - i - 1] + x;
      mp1[num1[pos - i]]++;
      min1 = min(min1, num1[pos - i]);
      max1 = max(max1, num1[pos - i]);
    }
    for (i = pos + 1; i <= n; i++) {
      if (a[i] < m)
        x = -1;
      else
        x = 1;
      num2[i - pos] = num2[i - pos - 1] + x;
      mp2[num2[i - pos]]++;
      min2 = min(min2, num2[i - pos]);
      max2 = max(max2, num2[i - pos]);
    }
    ans = 1 + mp1[0] + mp2[0] + mp1[1] + mp2[1];
    cnt = 0;
    for (i = min2; i <= max2; i++) {
      cnt += (long long)((long long)mp1[-i] + mp1[-i + 1]) * mp2[i];
    }
    for (i = min1; i <= max1; i++) {
      cnt += (long long)((long long)mp2[-i] + mp2[-i + 1]) * mp1[i];
    }
    printf("%lld\n", ans + cnt / 2);
  }
  return 0;
}

```