#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
struct Segment {
  vector<int> tree;
  int size;
  Segment() {}
  Segment(int N) : size(N + 1) { tree.resize(4 * (N + 1), 0); }
  int update(int idx, int val, int node, int nl, int nr) {
    if (idx < nl || idx > nr) return tree[node];
    if (nl == nr) return tree[node] += val;
    int mid = (nl + nr) >> 1;
    return tree[node] = update(idx, val, node * 2, nl, mid) +
                        update(idx, val, node * 2 + 1, mid + 1, nr);
  }
  void update(int idx, int val) { update(idx, val, 1, 0, size - 1); }
  int query(int val, int node, int nl, int nr) {
    if (nl == nr) return nl;
    int mid = (nl + nr) >> 1;
    if (tree[node * 2] >= val)
      return query(val, node * 2, nl, mid);
    else
      return query(val - tree[node * 2], node * 2 + 1, mid + 1, nr);
  }
  int query(int val) { return query(val, 1, 0, size - 1); }
};
int n, m, idx;
int arr[MAXN];
int l[MAXN], r[MAXN];
map<int, int> odd, even;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == m) idx = i;
  }
  for (int i = idx - 1; i >= 0; i--) {
    l[i] = l[i + 1];
    if (arr[i] < m)
      l[i]--;
    else
      l[i]++;
  }
  for (int i = idx; i < n; i++) {
    r[i] = r[i - 1];
    if (arr[i] > m)
      r[i]++;
    else if (arr[i] < m)
      r[i]--;
    if (i & 1)
      odd[r[i]]++;
    else
      even[r[i]]++;
  }
  long long ans = 0;
  l[idx] = 0;
  for (int i = idx; i >= 0; i--) {
    if (i & 1)
      ans += (long long)even[-l[i] + 1] + odd[-l[i]];
    else
      ans += (long long)odd[-l[i] + 1] + even[-l[i]];
  }
  printf("%lld\n", ans);
  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 int MAXN = 2e5 + 5;
struct Segment {
  vector<int> tree;
  int size;
  Segment() {}
  Segment(int N) : size(N + 1) { tree.resize(4 * (N + 1), 0); }
  int update(int idx, int val, int node, int nl, int nr) {
    if (idx < nl || idx > nr) return tree[node];
    if (nl == nr) return tree[node] += val;
    int mid = (nl + nr) >> 1;
    return tree[node] = update(idx, val, node * 2, nl, mid) +
                        update(idx, val, node * 2 + 1, mid + 1, nr);
  }
  void update(int idx, int val) { update(idx, val, 1, 0, size - 1); }
  int query(int val, int node, int nl, int nr) {
    if (nl == nr) return nl;
    int mid = (nl + nr) >> 1;
    if (tree[node * 2] >= val)
      return query(val, node * 2, nl, mid);
    else
      return query(val - tree[node * 2], node * 2 + 1, mid + 1, nr);
  }
  int query(int val) { return query(val, 1, 0, size - 1); }
};
int n, m, idx;
int arr[MAXN];
int l[MAXN], r[MAXN];
map<int, int> odd, even;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    if (arr[i] == m) idx = i;
  }
  for (int i = idx - 1; i >= 0; i--) {
    l[i] = l[i + 1];
    if (arr[i] < m)
      l[i]--;
    else
      l[i]++;
  }
  for (int i = idx; i < n; i++) {
    r[i] = r[i - 1];
    if (arr[i] > m)
      r[i]++;
    else if (arr[i] < m)
      r[i]--;
    if (i & 1)
      odd[r[i]]++;
    else
      even[r[i]]++;
  }
  long long ans = 0;
  l[idx] = 0;
  for (int i = idx; i >= 0; i--) {
    if (i & 1)
      ans += (long long)even[-l[i] + 1] + odd[-l[i]];
    else
      ans += (long long)odd[-l[i] + 1] + even[-l[i]];
  }
  printf("%lld\n", ans);
  return 0;
}

```