#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1000006;
long long num[maxn], c[maxn], n, m;
inline void update(long long x) {
  for (long long i = x; i <= n * 2 + 1; i += (i & (-i))) c[i] += 1;
}
inline long long query(long long x) {
  long long ans = 0;
  for (long long i = x; i > 0; i -= i & (-i)) ans += c[i];
  return ans;
}
long long sum[maxn], bg[maxn], sm[maxn];
inline long long solve(long long k) {
  long long ans = 0;
  memset(c, 0, sizeof(c));
  memset(sum, 0, sizeof(sum));
  memset(bg, 0, sizeof(bg));
  memset(sm, 0, sizeof(sum));
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 1; i <= n; ++i) {
    if (num[i] > k)
      cnt1++, sm[i] = sm[i - 1], bg[i] = cnt1;
    else
      cnt2++, bg[i] = bg[i - 1], sm[i] = cnt2;
  }
  for (long long i = 1; i <= n; ++i) sum[i] = sm[i] - bg[i];
  update(n + 1);
  for (long long i = 1; i <= n; ++i) {
    ans += query(n + 1 + sum[i]);
    update(n + 1 + sum[i]);
  }
  return ans;
}
signed main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) scanf("%lld", &num[i]);
  printf("%lld", solve(m) - solve(m - 1));
  return 0;
}

### Prompt

Please formulate 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>
using namespace std;
const long long maxn = 1000006;
long long num[maxn], c[maxn], n, m;
inline void update(long long x) {
  for (long long i = x; i <= n * 2 + 1; i += (i & (-i))) c[i] += 1;
}
inline long long query(long long x) {
  long long ans = 0;
  for (long long i = x; i > 0; i -= i & (-i)) ans += c[i];
  return ans;
}
long long sum[maxn], bg[maxn], sm[maxn];
inline long long solve(long long k) {
  long long ans = 0;
  memset(c, 0, sizeof(c));
  memset(sum, 0, sizeof(sum));
  memset(bg, 0, sizeof(bg));
  memset(sm, 0, sizeof(sum));
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 1; i <= n; ++i) {
    if (num[i] > k)
      cnt1++, sm[i] = sm[i - 1], bg[i] = cnt1;
    else
      cnt2++, bg[i] = bg[i - 1], sm[i] = cnt2;
  }
  for (long long i = 1; i <= n; ++i) sum[i] = sm[i] - bg[i];
  update(n + 1);
  for (long long i = 1; i <= n; ++i) {
    ans += query(n + 1 + sum[i]);
    update(n + 1 + sum[i]);
  }
  return ans;
}
signed main() {
  cin >> n >> m;
  for (long long i = 1; i <= n; ++i) scanf("%lld", &num[i]);
  printf("%lld", solve(m) - solve(m - 1));
  return 0;
}

```