#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
struct segtree {
  vector<long long> t;
  segtree(int n) {
    int sz = 1;
    while (sz < n) sz *= 2;
    t.resize(2 * sz, 0);
  }
  long long upd(int v, int vl, int vr, int p, long long x) {
    if (vl > p || vr <= p) {
      return t[v];
    }
    if (vr - vl == 1) {
      return t[v] += x;
    }
    int vm = (vl + vr) / 2;
    return t[v] = upd(v * 2, vl, vm, p, x) + upd(v * 2 + 1, vm, vr, p, x);
  }
  void upd(int p, long long x) { upd(1, 0, t.size() / 2, p, x); }
  long long get(int v, int vl, int vr, int l, int r) {
    if (vl >= r || vr <= l) {
      return 0;
    }
    if (l <= vl && vr <= r) {
      return t[v];
    }
    int vm = (vl + vr) / 2;
    return get(v * 2, vl, vm, l, r) + get(v * 2 + 1, vm, vr, l, r);
  }
  long long get(int l, int r) { return get(1, 0, t.size() / 2, l, r + 1); }
};
vector<int> a;
long long solve(int m) {
  segtree t(2 * N);
  long long res = 0;
  int k = N - 5;
  t.upd(k, 1);
  for (int i = 0; i < a.size(); i++) {
    if (a[i] >= m)
      k++;
    else
      k--;
    res += t.get(0, k - 1);
    t.upd(k, 1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
  }
  cout << solve(m) - solve(m + 1);
  return 0;
}

### Prompt

Create a solution in cpp for 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 N = 2e5 + 5;
struct segtree {
  vector<long long> t;
  segtree(int n) {
    int sz = 1;
    while (sz < n) sz *= 2;
    t.resize(2 * sz, 0);
  }
  long long upd(int v, int vl, int vr, int p, long long x) {
    if (vl > p || vr <= p) {
      return t[v];
    }
    if (vr - vl == 1) {
      return t[v] += x;
    }
    int vm = (vl + vr) / 2;
    return t[v] = upd(v * 2, vl, vm, p, x) + upd(v * 2 + 1, vm, vr, p, x);
  }
  void upd(int p, long long x) { upd(1, 0, t.size() / 2, p, x); }
  long long get(int v, int vl, int vr, int l, int r) {
    if (vl >= r || vr <= l) {
      return 0;
    }
    if (l <= vl && vr <= r) {
      return t[v];
    }
    int vm = (vl + vr) / 2;
    return get(v * 2, vl, vm, l, r) + get(v * 2 + 1, vm, vr, l, r);
  }
  long long get(int l, int r) { return get(1, 0, t.size() / 2, l, r + 1); }
};
vector<int> a;
long long solve(int m) {
  segtree t(2 * N);
  long long res = 0;
  int k = N - 5;
  t.upd(k, 1);
  for (int i = 0; i < a.size(); i++) {
    if (a[i] >= m)
      k++;
    else
      k--;
    res += t.get(0, k - 1);
    t.upd(k, 1);
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a.push_back(x);
  }
  cout << solve(m) - solve(m + 1);
  return 0;
}

```