#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long inf = 3e18 + 5;
long long add(long long a, long long b) { return (a += b) < mod ? a : a - mod; }
long long mul(long long a, long long b) { return 1LL * a * b % mod; }
long long sub(long long a, long long b) { return (a -= b) < 0 ? a + mod : a; }
long long ctz(long long x) { return __builtin_ctz(x); }
long long clz(long long x) { return __builtin_clz(x); }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n);
  long long aa = 0, ind = 0;
  for (long long& x : v) {
    cin >> x;
    if (x == m) {
      ind = aa;
      aa++;
      continue;
    }
    if (x < m)
      x = -1;
    else
      x = 1;
    aa++;
  }
  map<long long, long long> suf;
  suf[0] = 1;
  long long cum = 0;
  for (long long i = ind + 1; i < n; i++) {
    cum += v[i];
    suf[cum]++;
  }
  long long res = 0;
  long long pref = 0;
  for (long long i = ind; i >= 0; i--) {
    pref += (i == ind ? 0 : v[i]);
    res += suf[-pref];
    res += suf[1 - pref];
  }
  cout << res;
}

### 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;
const long long mod = 1e9 + 7;
const long long inf = 3e18 + 5;
long long add(long long a, long long b) { return (a += b) < mod ? a : a - mod; }
long long mul(long long a, long long b) { return 1LL * a * b % mod; }
long long sub(long long a, long long b) { return (a -= b) < 0 ? a + mod : a; }
long long ctz(long long x) { return __builtin_ctz(x); }
long long clz(long long x) { return __builtin_clz(x); }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n);
  long long aa = 0, ind = 0;
  for (long long& x : v) {
    cin >> x;
    if (x == m) {
      ind = aa;
      aa++;
      continue;
    }
    if (x < m)
      x = -1;
    else
      x = 1;
    aa++;
  }
  map<long long, long long> suf;
  suf[0] = 1;
  long long cum = 0;
  for (long long i = ind + 1; i < n; i++) {
    cum += v[i];
    suf[cum]++;
  }
  long long res = 0;
  long long pref = 0;
  for (long long i = ind; i >= 0; i--) {
    pref += (i == ind ? 0 : v[i]);
    res += suf[-pref];
    res += suf[1 - pref];
  }
  cout << res;
}

```