#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
struct BIT {
 private:
  V<T> arr;
  int n;

 public:
  BIT(int siz) {
    n = siz;
    arr.assign(n + 1, 0);
  }
  void add(T a, T v) {
    for (T x = a; x <= n; x += (x & -x)) {
      arr[x] += v;
    }
  }
  T sum(T a) {
    T ret = 0;
    for (T x = a; x > 0; x -= (x & -x)) ret += arr[x];
    return ret;
  }
  T lower_bound(T w) {
    if (w <= 0) return 0;
    T x, r = 1;
    while (r < n) r <<= 1;
    for (T k = r; k > 0; k >>= 1) {
      if (x + k <= n && arr[x + r] < w) {
        w -= arr[x + r];
        x += k;
      }
    }
    return x + 1;
  }
  T get(T x) {
    if (x <= 0) return 0;
    T N = 1;
    while (N * 2 <= n) N *= 2;
    T v = 0;
    for (T i = N; i > 0; i /= 2) {
      if (v + i <= n && arr[v + i] < x) {
        x -= arr[v + i];
        v += i;
      }
    }
    return v + 1;
  }
  T query(int l, int r) { return sum(r) - sum(l); }
};
int n, m;
V<ll> a(n);
V<int> d;
ll solve(ll x) {
  d.assign(n, 0);
  BIT<ll> bit(2 * n + 10);
  for (int i = 0; i < n; i++) {
    if (a[i] >= x)
      d[i] = 1;
    else
      d[i] = -1;
  }
  int now = 0;
  ll res = 0;
  for (int i = 0; i < n; i++) {
    bit.add(n + now, 1);
    now += d[i];
    res += bit.sum(n + now - 1);
  }
  return res;
}
int main() {
  cin >> n >> m;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << solve(m) - solve(m + 1) << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
template <class T>
using V = vector<T>;
const ll inf = (1e18);
const ll mod = 1000000007;
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll c, ll d) { return c / GCD(c, d) * d; }
struct __INIT {
  __INIT() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(15);
  }
} __init;
template <class T>
bool chmax(T &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T>
struct BIT {
 private:
  V<T> arr;
  int n;

 public:
  BIT(int siz) {
    n = siz;
    arr.assign(n + 1, 0);
  }
  void add(T a, T v) {
    for (T x = a; x <= n; x += (x & -x)) {
      arr[x] += v;
    }
  }
  T sum(T a) {
    T ret = 0;
    for (T x = a; x > 0; x -= (x & -x)) ret += arr[x];
    return ret;
  }
  T lower_bound(T w) {
    if (w <= 0) return 0;
    T x, r = 1;
    while (r < n) r <<= 1;
    for (T k = r; k > 0; k >>= 1) {
      if (x + k <= n && arr[x + r] < w) {
        w -= arr[x + r];
        x += k;
      }
    }
    return x + 1;
  }
  T get(T x) {
    if (x <= 0) return 0;
    T N = 1;
    while (N * 2 <= n) N *= 2;
    T v = 0;
    for (T i = N; i > 0; i /= 2) {
      if (v + i <= n && arr[v + i] < x) {
        x -= arr[v + i];
        v += i;
      }
    }
    return v + 1;
  }
  T query(int l, int r) { return sum(r) - sum(l); }
};
int n, m;
V<ll> a(n);
V<int> d;
ll solve(ll x) {
  d.assign(n, 0);
  BIT<ll> bit(2 * n + 10);
  for (int i = 0; i < n; i++) {
    if (a[i] >= x)
      d[i] = 1;
    else
      d[i] = -1;
  }
  int now = 0;
  ll res = 0;
  for (int i = 0; i < n; i++) {
    bit.add(n + now, 1);
    now += d[i];
    res += bit.sum(n + now - 1);
  }
  return res;
}
int main() {
  cin >> n >> m;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  cout << solve(m) - solve(m + 1) << "\n";
}

```