#include <bits/stdc++.h>
using namespace std;
string itosm(long long x) {
  if (x == 0) return "0";
  string res = "";
  while (x > 0) {
    res += ((x % 10) + '0');
    x /= 10;
  }
  reverse(res.begin(), res.end());
  return res;
}
long long stoim(string str) {
  long long res = 0;
  int p = 0;
  if (str[0] == '-') p++;
  for (int i = p; i < str.length(); i++) {
    res *= 10;
    res += (str[i] - '0');
  }
  return res;
}
const long long infll = 1e18 + 3;
const int inf = 1009000999;
const long double eps = 1e-7;
const int maxn = 1e6 + 1146;
const int baseint = 1000200013;
const long long basell = 1e18 + 3;
const long double PI = acos(-1.0);
const long long mod = 998244353;
long long binpow(long long x, long long st) {
  if (st == 0) return 1;
  if (st & 1) return binpow(x, st - 1) * x % mod;
  long long res = binpow(x, st >> 1);
  return res * res % mod;
}
int a[maxn];
int s[maxn];
int n, m;
long long f(int x) {
  long long res = 0;
  for (int i = 0; i < maxn; i++) s[i] = 0;
  int sum = n;
  s[sum] = 1;
  long long add = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] < x)
      sum--, add -= s[sum];
    else
      add += s[sum], sum++;
    res += add;
    s[sum]++;
  }
  return res;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << f(m) - f(m + 1);
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  ;
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
string itosm(long long x) {
  if (x == 0) return "0";
  string res = "";
  while (x > 0) {
    res += ((x % 10) + '0');
    x /= 10;
  }
  reverse(res.begin(), res.end());
  return res;
}
long long stoim(string str) {
  long long res = 0;
  int p = 0;
  if (str[0] == '-') p++;
  for (int i = p; i < str.length(); i++) {
    res *= 10;
    res += (str[i] - '0');
  }
  return res;
}
const long long infll = 1e18 + 3;
const int inf = 1009000999;
const long double eps = 1e-7;
const int maxn = 1e6 + 1146;
const int baseint = 1000200013;
const long long basell = 1e18 + 3;
const long double PI = acos(-1.0);
const long long mod = 998244353;
long long binpow(long long x, long long st) {
  if (st == 0) return 1;
  if (st & 1) return binpow(x, st - 1) * x % mod;
  long long res = binpow(x, st >> 1);
  return res * res % mod;
}
int a[maxn];
int s[maxn];
int n, m;
long long f(int x) {
  long long res = 0;
  for (int i = 0; i < maxn; i++) s[i] = 0;
  int sum = n;
  s[sum] = 1;
  long long add = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] < x)
      sum--, add -= s[sum];
    else
      add += s[sum], sum++;
    res += add;
    s[sum]++;
  }
  return res;
}
void solve() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << f(m) - f(m + 1);
}
int main() {
  srand(time(0));
  ios_base::sync_with_stdio(0);
  ;
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}

```