#include <bits/stdc++.h>
using namespace std;
template <class T>
struct rge {
  T b, e;
};
template <class T>
rge<T> range(T i, T j) {
  return rge<T>{i, j};
};
template <class A, class B>
ostream &operator<<(ostream &os, pair<A, B> p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, rge<T> d) {
  os << "[";
  for (auto it = d.b; it != d.e; it++) os << (it == d.b ? "" : ", ") << *it;
  return os << "]";
}
const int N = 2e5 + 100;
const int inf = 1e9;
template <class T>
struct fenwick {
  vector<T> fenw;
  fenwick(int _n = 0) { fenw.resize(_n + 5, 0); }
  void update(int sx, T v) {
    for (int x = (sx + N); x <= ((int)(fenw).size() - 1); x += (+(x & -x))) {
      fenw[x] += v;
    }
  }
  T query(int sx) {
    T v = 0;
    for (int x = (sx + N); x >= (1); x += (-(x & -x))) {
      v += fenw[x];
    }
    return v;
  }
};
int a[N];
int n, med;
long long solve(int med) {
  fenwick<long long> cnt(2 * N);
  cnt.update(0, +1);
  long long tot = 0, more = 0;
  for (int i = (0); i <= ((n)-1); i += (+1)) {
    if (a[i] <= med) {
      more--;
    } else {
      more++;
    }
    tot += cnt.query(-more);
    cnt.update(-more, +1);
  }
  return tot;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(15);
  while (cin >> n >> med) {
    for (int i = (0); i <= ((n)-1); i += (+1)) {
      cin >> a[i];
    }
    cout << solve(med) - solve(med - 1) << '\n';
  }
  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;
template <class T>
struct rge {
  T b, e;
};
template <class T>
rge<T> range(T i, T j) {
  return rge<T>{i, j};
};
template <class A, class B>
ostream &operator<<(ostream &os, pair<A, B> p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream &operator<<(ostream &os, rge<T> d) {
  os << "[";
  for (auto it = d.b; it != d.e; it++) os << (it == d.b ? "" : ", ") << *it;
  return os << "]";
}
const int N = 2e5 + 100;
const int inf = 1e9;
template <class T>
struct fenwick {
  vector<T> fenw;
  fenwick(int _n = 0) { fenw.resize(_n + 5, 0); }
  void update(int sx, T v) {
    for (int x = (sx + N); x <= ((int)(fenw).size() - 1); x += (+(x & -x))) {
      fenw[x] += v;
    }
  }
  T query(int sx) {
    T v = 0;
    for (int x = (sx + N); x >= (1); x += (-(x & -x))) {
      v += fenw[x];
    }
    return v;
  }
};
int a[N];
int n, med;
long long solve(int med) {
  fenwick<long long> cnt(2 * N);
  cnt.update(0, +1);
  long long tot = 0, more = 0;
  for (int i = (0); i <= ((n)-1); i += (+1)) {
    if (a[i] <= med) {
      more--;
    } else {
      more++;
    }
    tot += cnt.query(-more);
    cnt.update(-more, +1);
  }
  return tot;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cout.precision(15);
  while (cin >> n >> med) {
    for (int i = (0); i <= ((n)-1); i += (+1)) {
      cin >> a[i];
    }
    cout << solve(med) - solve(med - 1) << '\n';
  }
  return 0;
}

```