#include <bits/stdc++.h>
using namespace std;
const int maxN = 200000 + 9;
int a[maxN], pre[maxN];
int bit[2 * maxN];
int n, m;
int lowbit(int x) { return x & (-x); }
void modify(int pos, int add) {
  while (pos <= 2 * n) {
    bit[pos] += add;
    pos += lowbit(pos);
  }
}
long long query(int pos) {
  long long res = 0;
  while (pos > 0) {
    res += bit[pos];
    pos -= lowbit(pos);
  }
  return res;
}
long long fun(int x) {
  long long res = 0;
  memset(bit, 0, sizeof bit);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i] <= x ? 1 : -1);
  modify(n + 1, 1);
  for (int i = 1; i <= n; i++) {
    res += query(pre[i] + n + 1);
    modify(pre[i] + n + 1, 1);
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  printf("%lld\n", fun(m) - fun(m - 1));
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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 = 200000 + 9;
int a[maxN], pre[maxN];
int bit[2 * maxN];
int n, m;
int lowbit(int x) { return x & (-x); }
void modify(int pos, int add) {
  while (pos <= 2 * n) {
    bit[pos] += add;
    pos += lowbit(pos);
  }
}
long long query(int pos) {
  long long res = 0;
  while (pos > 0) {
    res += bit[pos];
    pos -= lowbit(pos);
  }
  return res;
}
long long fun(int x) {
  long long res = 0;
  memset(bit, 0, sizeof bit);
  for (int i = 1; i <= n; i++) pre[i] = pre[i - 1] + (a[i] <= x ? 1 : -1);
  modify(n + 1, 1);
  for (int i = 1; i <= n; i++) {
    res += query(pre[i] + n + 1);
    modify(pre[i] + n + 1, 1);
  }
  return res;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  printf("%lld\n", fun(m) - fun(m - 1));
  return 0;
}

```