#include <bits/stdc++.h>
using namespace std;
const long long int logx = 30;
const long long int N = 3e5 + 5;
const long long int M = 3e5 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265358979323846;
inline long long int mul(long long int a, long long int b) {
  return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (c > mod) c -= mod;
  return c;
}
template <typename T>
T power(T x, T y, long long int m) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
map<long long int, long long int> cnt;
long long int n, m, a[N], pre[N], seg[4 * N];
void update(long long int low, long long int high, long long int qpos,
            long long int pos) {
  if (low == high) {
    seg[pos]++;
    return;
  } else {
    long long int mid = (low + high) / 2;
    if (low <= qpos && qpos <= mid) {
      update(low, mid, qpos, 2 * pos + 1);
    } else {
      update(mid + 1, high, qpos, 2 * pos + 2);
    }
    seg[pos] = seg[2 * pos + 1] + seg[2 * pos + 2];
  }
}
long long int query(long long int low, long long int high, long long int qlow,
                    long long int qhigh, long long int pos) {
  if (qlow > high || qhigh < low) {
    return 0;
  }
  if (qlow <= low && qhigh >= high) {
    return seg[pos];
  }
  long long int mid = (low + high) / 2;
  return query(low, mid, qlow, qhigh, 2 * pos + 1) +
         query(mid + 1, high, qlow, qhigh, 2 * pos + 2);
}
long long int F(long long int cur) {
  memset(seg, 0, sizeof(seg));
  long long int sum = 0, ans = 0;
  vector<long long int> pending;
  update(0, 2 * n, n, 0);
  for (long long int i = 0; i < n; i++) {
    if (a[i] >= cur) {
      for (auto j : pending) {
        update(0, 2 * n, j + n, 0);
      }
      pending.clear();
    }
    sum += ((a[i] >= cur) - (a[i] < cur));
    ans += query(0, 2 * n, 0, sum + n - 1, 0);
    pending.push_back(sum);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << F(m) - F(m + 1) << endl;
}

### Prompt

In cpp, your task is to solve 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 int logx = 30;
const long long int N = 3e5 + 5;
const long long int M = 3e5 + 5;
const long long int mod = 1e9 + 7;
const long long int INF = 1e18 + 5;
const double PI = 3.14159265358979323846;
inline long long int mul(long long int a, long long int b) {
  return (a * 1ll * b) % mod;
}
inline long long int sub(long long int a, long long int b) {
  long long int c = a - b;
  if (c < 0) c += mod;
  return c;
}
inline long long int add(long long int a, long long int b) {
  long long int c = a + b;
  if (c > mod) c -= mod;
  return c;
}
template <typename T>
T power(T x, T y, long long int m) {
  T ans = 1;
  while (y > 0) {
    if (y & 1LL) ans = (ans * x) % m;
    y >>= 1ll;
    x = (x * x) % m;
  }
  return ans % m;
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
map<long long int, long long int> cnt;
long long int n, m, a[N], pre[N], seg[4 * N];
void update(long long int low, long long int high, long long int qpos,
            long long int pos) {
  if (low == high) {
    seg[pos]++;
    return;
  } else {
    long long int mid = (low + high) / 2;
    if (low <= qpos && qpos <= mid) {
      update(low, mid, qpos, 2 * pos + 1);
    } else {
      update(mid + 1, high, qpos, 2 * pos + 2);
    }
    seg[pos] = seg[2 * pos + 1] + seg[2 * pos + 2];
  }
}
long long int query(long long int low, long long int high, long long int qlow,
                    long long int qhigh, long long int pos) {
  if (qlow > high || qhigh < low) {
    return 0;
  }
  if (qlow <= low && qhigh >= high) {
    return seg[pos];
  }
  long long int mid = (low + high) / 2;
  return query(low, mid, qlow, qhigh, 2 * pos + 1) +
         query(mid + 1, high, qlow, qhigh, 2 * pos + 2);
}
long long int F(long long int cur) {
  memset(seg, 0, sizeof(seg));
  long long int sum = 0, ans = 0;
  vector<long long int> pending;
  update(0, 2 * n, n, 0);
  for (long long int i = 0; i < n; i++) {
    if (a[i] >= cur) {
      for (auto j : pending) {
        update(0, 2 * n, j + n, 0);
      }
      pending.clear();
    }
    sum += ((a[i] >= cur) - (a[i] < cur));
    ans += query(0, 2 * n, 0, sum + n - 1, 0);
    pending.push_back(sum);
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  cin >> n >> m;
  for (long long int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << F(m) - F(m + 1) << endl;
}

```