#include <bits/stdc++.h>
using namespace std;
bool down[300000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, acu = 0, foo, saw;
  bool seen = false;
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) {
    cin >> foo;
    if (foo == m) saw = i;
    down[i] = (foo > m);
  }
  map<pair<long long, long long>, long long> arriba, abajo;
  long long l = 0, mo = 0;
  for (long long i = saw - 1; i >= 0; --i) {
    l += (down[i] == 0);
    mo += (down[i] == 1);
    abajo[make_pair(l - min(l, mo), mo - min(l, mo))]++;
  }
  l = 0;
  mo = 0;
  for (long long i = saw + 1; i < n; ++i) {
    l += (down[i] == 0);
    mo += (down[i] == 1);
    arriba[make_pair(l - min(l, mo), mo - min(l, mo))]++;
  }
  long long rta = 0;
  for (auto i : arriba) {
    if (abajo[make_pair(i.first.second, i.first.first)] > 0)
      rta += i.second * abajo[make_pair(i.first.second, i.first.first)];
    if (abajo[make_pair(i.first.second, i.first.first + 1)] > 0)
      rta += i.second *
             abajo[make_pair(
                 i.first.second - min(i.first.second, i.first.first + 1),
                 i.first.first + 1 - min(i.first.second, i.first.first + 1))];
    if (i.first.second > 0 &&
        abajo[make_pair(i.first.second - 1, i.first.first)] > 0)
      rta += i.second * abajo[make_pair(i.first.second - 1, i.first.first)];
  }
  cout << rta + (long long)(abajo[make_pair(0, 0)] + abajo[make_pair(0, 1)] +
                            arriba[make_pair(0, 0)] + arriba[make_pair(0, 1)] +
                            1);
}

### 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;
bool down[300000];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, m, acu = 0, foo, saw;
  bool seen = false;
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) {
    cin >> foo;
    if (foo == m) saw = i;
    down[i] = (foo > m);
  }
  map<pair<long long, long long>, long long> arriba, abajo;
  long long l = 0, mo = 0;
  for (long long i = saw - 1; i >= 0; --i) {
    l += (down[i] == 0);
    mo += (down[i] == 1);
    abajo[make_pair(l - min(l, mo), mo - min(l, mo))]++;
  }
  l = 0;
  mo = 0;
  for (long long i = saw + 1; i < n; ++i) {
    l += (down[i] == 0);
    mo += (down[i] == 1);
    arriba[make_pair(l - min(l, mo), mo - min(l, mo))]++;
  }
  long long rta = 0;
  for (auto i : arriba) {
    if (abajo[make_pair(i.first.second, i.first.first)] > 0)
      rta += i.second * abajo[make_pair(i.first.second, i.first.first)];
    if (abajo[make_pair(i.first.second, i.first.first + 1)] > 0)
      rta += i.second *
             abajo[make_pair(
                 i.first.second - min(i.first.second, i.first.first + 1),
                 i.first.first + 1 - min(i.first.second, i.first.first + 1))];
    if (i.first.second > 0 &&
        abajo[make_pair(i.first.second - 1, i.first.first)] > 0)
      rta += i.second * abajo[make_pair(i.first.second - 1, i.first.first)];
  }
  cout << rta + (long long)(abajo[make_pair(0, 0)] + abajo[make_pair(0, 1)] +
                            arriba[make_pair(0, 0)] + arriba[make_pair(0, 1)] +
                            1);
}

```