#include <bits/stdc++.h>
static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;
template <class T>
constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;
template <class T>
class BIT {
  vector<T> bit;

 public:
  BIT(int n) : bit(vector<T>(n + 1, 0)) {}
  T sum(int k) {
    T ret = 0;
    for (++k; k > 0; k -= (k & -k)) ret += bit[k];
    return ret;
  }
  void add(int k, T x) {
    for (++k; k < bit.size(); k += (k & -k)) bit[k] += x;
  }
};
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  for (auto &&i : v) scanf("%d", &i);
  auto f = [&](int x) -> ll {
    ll ret = 0;
    BIT<int> S(2 * n + 1);
    vector<int> T(n + 1);
    for (int i = 0; i < n; ++i) T[i + 1] = T[i] + (v[i] <= x ? 1 : -1);
    int mn = *min_element(T.begin(), T.end());
    for (auto &&i : T) {
      ret += S.sum(i - mn);
      S.add(i - mn, 1);
    }
    return ret;
  };
  cout << f(m) - f(m - 1) << "\n";
  return 0;
}

### Prompt

Generate a cpp solution to 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>
static const int MOD = 1000000007;
using ll = long long;
using u32 = unsigned;
using u64 = unsigned long long;
using namespace std;
template <class T>
constexpr T INF = ::numeric_limits<T>::max() / 32 * 15 + 208;
template <class T>
class BIT {
  vector<T> bit;

 public:
  BIT(int n) : bit(vector<T>(n + 1, 0)) {}
  T sum(int k) {
    T ret = 0;
    for (++k; k > 0; k -= (k & -k)) ret += bit[k];
    return ret;
  }
  void add(int k, T x) {
    for (++k; k < bit.size(); k += (k & -k)) bit[k] += x;
  }
};
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> v(n);
  for (auto &&i : v) scanf("%d", &i);
  auto f = [&](int x) -> ll {
    ll ret = 0;
    BIT<int> S(2 * n + 1);
    vector<int> T(n + 1);
    for (int i = 0; i < n; ++i) T[i + 1] = T[i] + (v[i] <= x ? 1 : -1);
    int mn = *min_element(T.begin(), T.end());
    for (auto &&i : T) {
      ret += S.sum(i - mn);
      S.add(i - mn, 1);
    }
    return ret;
  };
  cout << f(m) - f(m - 1) << "\n";
  return 0;
}

```