#include <bits/stdc++.h>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[400005];
long long b[400005], tr[400005];
void update(long long x, long long val) {
  while (x < 400005) {
    tr[x] += val;
    x += x & (-x);
  }
}
long long query(long long x) {
  long long s = 0;
  while (x > 0) {
    s += tr[x];
    x -= x & (-x);
  }
  return s;
}
long long get(long long x, long long n) {
  memset(tr, 0, sizeof(tr));
  long long i, j = 0;
  long long m = 200001;
  long long s = 0;
  update(m, 1);
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    if (a[i] < x)
      s--;
    else
      s++;
    ans += query(s + m - 1);
    update(s + m, 1);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc, i, j, n, m, l, r;
  long long cn = 0;
  while (cin >> n >> m) {
    long long pos, ans = 0;
    for (i = 1; i <= n; i++) cin >> a[i];
    cout << get(m, n) - get(m + 1, n) << endl;
  }
  return 0;
}

### Prompt

Please formulate 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>
int dr[] = {2, 2, -2, -2, 1, -1, 1, -1};
int dc[] = {1, -1, 1, -1, 2, 2, -2, -2};
int dr1[] = {0, 0, 1, 1, 1, -1, -1, -1};
int dc1[] = {1, -1, 1, 0, -1, 0, 1, -1};
int dr2[] = {0, 0, 1, -1};
int dc2[] = {1, -1, 0, 0};
using namespace std;
long long a[400005];
long long b[400005], tr[400005];
void update(long long x, long long val) {
  while (x < 400005) {
    tr[x] += val;
    x += x & (-x);
  }
}
long long query(long long x) {
  long long s = 0;
  while (x > 0) {
    s += tr[x];
    x -= x & (-x);
  }
  return s;
}
long long get(long long x, long long n) {
  memset(tr, 0, sizeof(tr));
  long long i, j = 0;
  long long m = 200001;
  long long s = 0;
  update(m, 1);
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    if (a[i] < x)
      s--;
    else
      s++;
    ans += query(s + m - 1);
    update(s + m, 1);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc, i, j, n, m, l, r;
  long long cn = 0;
  while (cin >> n >> m) {
    long long pos, ans = 0;
    for (i = 1; i <= n; i++) cin >> a[i];
    cout << get(m, n) - get(m + 1, n) << endl;
  }
  return 0;
}

```