#include <bits/stdc++.h>
using namespace std;
template <typename T>
struct BIT {
  int n;
  vector<T> bit;
  BIT(int n_ = 0) : n(n_), bit(n_ + 1) {}
  T sum(int i) {
    T ans = 0;
    for (; i > 0; i -= i & -i) ans += bit[i];
    return ans;
  }
  void add(int i, T a) {
    if (i == 0) return;
    for (; i <= n; i += i & -i) bit[i] += a;
  }
  int lower_bound(T k) {
    if (k <= 0) return 0;
    int ret = 0, i = 1;
    while ((i << 1) <= n) i <<= 1;
    for (; i; i >>= 1)
      if (ret + i <= n && bit[ret + i] < k) k -= bit[ret += i];
    return ret + 1;
  }
};
int N, M;
int A[2 << 17];
int B[2 << 17];
long long f(int M) {
  vector<int> vs;
  vs.push_back(0);
  for (int i = 0; i < N; i++) {
    int a = B[i];
    a = a <= M ? 1 : -1;
    A[i + 1] = a + A[i];
    vs.push_back(A[i + 1]);
  }
  sort(vs.begin(), vs.end());
  vs.erase(unique(vs.begin(), vs.end()), vs.end());
  BIT<int> P(vs.size());
  long long ans = 0;
  for (int i = 0; i <= N; i++) {
    int a = lower_bound(vs.begin(), vs.end(), A[i]) - vs.begin();
    ans += P.sum(a + 1);
    P.add(a + 1, 1);
  }
  return ans;
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++) scanf("%d", &B[i]);
  printf("%lld\n", f(M) - f(M - 1));
}

### 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 <typename T>
struct BIT {
  int n;
  vector<T> bit;
  BIT(int n_ = 0) : n(n_), bit(n_ + 1) {}
  T sum(int i) {
    T ans = 0;
    for (; i > 0; i -= i & -i) ans += bit[i];
    return ans;
  }
  void add(int i, T a) {
    if (i == 0) return;
    for (; i <= n; i += i & -i) bit[i] += a;
  }
  int lower_bound(T k) {
    if (k <= 0) return 0;
    int ret = 0, i = 1;
    while ((i << 1) <= n) i <<= 1;
    for (; i; i >>= 1)
      if (ret + i <= n && bit[ret + i] < k) k -= bit[ret += i];
    return ret + 1;
  }
};
int N, M;
int A[2 << 17];
int B[2 << 17];
long long f(int M) {
  vector<int> vs;
  vs.push_back(0);
  for (int i = 0; i < N; i++) {
    int a = B[i];
    a = a <= M ? 1 : -1;
    A[i + 1] = a + A[i];
    vs.push_back(A[i + 1]);
  }
  sort(vs.begin(), vs.end());
  vs.erase(unique(vs.begin(), vs.end()), vs.end());
  BIT<int> P(vs.size());
  long long ans = 0;
  for (int i = 0; i <= N; i++) {
    int a = lower_bound(vs.begin(), vs.end(), A[i]) - vs.begin();
    ans += P.sum(a + 1);
    P.add(a + 1, 1);
  }
  return ans;
}
int main() {
  scanf("%d%d", &N, &M);
  for (int i = 0; i < N; i++) scanf("%d", &B[i]);
  printf("%lld\n", f(M) - f(M - 1));
}

```