#include <bits/stdc++.h>
using namespace std;
const int MAXN = 400010;
const int OFFSET = 200002;
int n, m;
int a[MAXN], b[MAXN];
namespace BIT {
const int n = OFFSET + OFFSET;
long long c0[MAXN], c1[MAXN];
void clear() {
  memset(c0, 0, sizeof(c0));
  memset(c1, 0, sizeof(c1));
}
void add(long long *c, int x, long long k) {
  for (; x <= n; x += x & -x) c[x] += k;
}
long long sum(long long *c, int x) {
  long long r = 0;
  for (; x; x -= x & -x) r += c[x];
  return r;
}
void add(int l, int r, long long k) {
  add(c0, l, k);
  add(c0, r + 1, -k);
  add(c1, l, (l - 1) * k);
  add(c1, r + 1, -r * k);
}
long long sum(int l, int r) {
  return sum(c0, r) * r - sum(c1, r) - sum(c0, l - 1) * (l - 1) +
         sum(c1, l - 1);
}
}  // namespace BIT
long long calc(int x) {
  BIT::clear();
  for (int i = 1; i <= n; i++) {
    b[i] = (a[i] >= x ? 1 : -1) + b[i - 1];
  }
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    BIT::add(b[i] + OFFSET, b[i] + OFFSET, 1);
    ans += BIT::sum(b[i - 1] + 1 + OFFSET, OFFSET + OFFSET);
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("%lld\n", calc(m) - calc(m + 1));
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 int MAXN = 400010;
const int OFFSET = 200002;
int n, m;
int a[MAXN], b[MAXN];
namespace BIT {
const int n = OFFSET + OFFSET;
long long c0[MAXN], c1[MAXN];
void clear() {
  memset(c0, 0, sizeof(c0));
  memset(c1, 0, sizeof(c1));
}
void add(long long *c, int x, long long k) {
  for (; x <= n; x += x & -x) c[x] += k;
}
long long sum(long long *c, int x) {
  long long r = 0;
  for (; x; x -= x & -x) r += c[x];
  return r;
}
void add(int l, int r, long long k) {
  add(c0, l, k);
  add(c0, r + 1, -k);
  add(c1, l, (l - 1) * k);
  add(c1, r + 1, -r * k);
}
long long sum(int l, int r) {
  return sum(c0, r) * r - sum(c1, r) - sum(c0, l - 1) * (l - 1) +
         sum(c1, l - 1);
}
}  // namespace BIT
long long calc(int x) {
  BIT::clear();
  for (int i = 1; i <= n; i++) {
    b[i] = (a[i] >= x ? 1 : -1) + b[i - 1];
  }
  long long ans = 0;
  for (int i = n; i >= 1; i--) {
    BIT::add(b[i] + OFFSET, b[i] + OFFSET, 1);
    ans += BIT::sum(b[i - 1] + 1 + OFFSET, OFFSET + OFFSET);
  }
  return ans;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  printf("%lld\n", calc(m) - calc(m + 1));
  return 0;
}

```