#include <bits/stdc++.h>
using namespace std;
void cline() { cout << '\n'; }
template <typename T, typename... V>
void cline(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ' ';
  cline(v...);
}
void cspc() { cout << ' '; }
template <typename T, typename... V>
void cspc(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ' ';
  cspc(v...);
}
const long long N = 2e5 + 4;
long long state[2 * N][2];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(state, 0, sizeof(state));
  long long n, m;
  cin >> n >> m;
  long long ind = -1;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (m == a[i]) ind = i;
  }
  long long oki = 0, cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] < m)
      oki--;
    else if (a[i] > m)
      oki++;
    if (i >= ind) state[N + oki][i & 1]++;
  }
  oki = 0;
  for (long long i = 0; i <= ind; ++i) {
    if (i % 2 == 1) {
      cnt += state[N + oki + 1][0];
      cnt += state[N + oki][1];
    } else {
      cnt += state[N + oki][0];
      cnt += state[N + oki + 1][1];
    }
    if (a[i] < m)
      oki--;
    else
      oki++;
  }
  cspc(cnt);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
void cline() { cout << '\n'; }
template <typename T, typename... V>
void cline(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ' ';
  cline(v...);
}
void cspc() { cout << ' '; }
template <typename T, typename... V>
void cspc(T t, V... v) {
  cout << t;
  if (sizeof...(v)) cout << ' ';
  cspc(v...);
}
const long long N = 2e5 + 4;
long long state[2 * N][2];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  memset(state, 0, sizeof(state));
  long long n, m;
  cin >> n >> m;
  long long ind = -1;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    if (m == a[i]) ind = i;
  }
  long long oki = 0, cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] < m)
      oki--;
    else if (a[i] > m)
      oki++;
    if (i >= ind) state[N + oki][i & 1]++;
  }
  oki = 0;
  for (long long i = 0; i <= ind; ++i) {
    if (i % 2 == 1) {
      cnt += state[N + oki + 1][0];
      cnt += state[N + oki][1];
    } else {
      cnt += state[N + oki][0];
      cnt += state[N + oki + 1][1];
    }
    if (a[i] < m)
      oki--;
    else
      oki++;
  }
  cspc(cnt);
  return 0;
}

```