#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
const int mod = 1e9 + 7;
int head[N];
int dis[N], ecnt;
int fa[N];
int cat[2005][2005];
long long gcd(long long a, long long b) {
  return a % b == 0 ? b : gcd(b, a % b);
}
long long qpow(long long base, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = (ans * base) % mod;
    n >>= 1;
    base = base * base % mod;
  }
  return ans;
}
struct Node {
  long long x, y;
} nd[N];
struct EDGE {
  int u, v, nxt;
  long long w;
} e[N];
bool cmp(EDGE a, EDGE b) { return a.w < b.w; }
void add_edge(int u, int v, long long w) {
  e[ecnt].u = u;
  e[ecnt].v = v;
  e[ecnt].w = w;
  e[ecnt].nxt = head[u];
  head[u] = ecnt++;
}
int fd(int x) { return -1 == fa[x] ? x : fa[x] = fd(fa[x]); }
long long m, n, x, k, y;
int c[N];
int lowbit(int x) { return x & (-x); }
void add(int x, int y) {
  while (x <= n) {
    c[x] += y;
    x += lowbit(x);
  }
}
int getsum(int x) {
  int res = 0;
  while (x > 0) {
    res += c[x];
    x -= lowbit(x);
  }
  return res;
}
int vis[N];
int p[N];
long long ans;
struct H {
  int p, s;
  friend bool operator<(H a, H b) { return a.p < b.p; }
} he[N];
bool cmp(int a, int b) { return a > b; }
priority_queue<int> q;
map<vector<int>, int> mp;
long long ar[N];
int br[N];
int pos[N];
bool cmpp(int a, int b) { return abs(ar[a] - ar[m]) < abs(ar[b] - ar[m]); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> ar[i];
      pos[ar[i]] = i;
    }
    int md = pos[m];
    int ct = 0;
    br[ct]++;
    for (int i = md + 1; i <= n; i++) {
      if (ar[i] > m)
        ct++;
      else
        ct--;
      br[ct]++;
    }
    ans = br[0] + br[1];
    ct = 0;
    for (int i = md - 1; i > 0; i--) {
      if (ar[i] < m)
        ct++;
      else
        ct--;
      ans += br[ct] + br[ct + 1];
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Generate 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 N = 1e6 + 10;
const int mod = 1e9 + 7;
int head[N];
int dis[N], ecnt;
int fa[N];
int cat[2005][2005];
long long gcd(long long a, long long b) {
  return a % b == 0 ? b : gcd(b, a % b);
}
long long qpow(long long base, long long n) {
  long long ans = 1;
  while (n) {
    if (n & 1) ans = (ans * base) % mod;
    n >>= 1;
    base = base * base % mod;
  }
  return ans;
}
struct Node {
  long long x, y;
} nd[N];
struct EDGE {
  int u, v, nxt;
  long long w;
} e[N];
bool cmp(EDGE a, EDGE b) { return a.w < b.w; }
void add_edge(int u, int v, long long w) {
  e[ecnt].u = u;
  e[ecnt].v = v;
  e[ecnt].w = w;
  e[ecnt].nxt = head[u];
  head[u] = ecnt++;
}
int fd(int x) { return -1 == fa[x] ? x : fa[x] = fd(fa[x]); }
long long m, n, x, k, y;
int c[N];
int lowbit(int x) { return x & (-x); }
void add(int x, int y) {
  while (x <= n) {
    c[x] += y;
    x += lowbit(x);
  }
}
int getsum(int x) {
  int res = 0;
  while (x > 0) {
    res += c[x];
    x -= lowbit(x);
  }
  return res;
}
int vis[N];
int p[N];
long long ans;
struct H {
  int p, s;
  friend bool operator<(H a, H b) { return a.p < b.p; }
} he[N];
bool cmp(int a, int b) { return a > b; }
priority_queue<int> q;
map<vector<int>, int> mp;
long long ar[N];
int br[N];
int pos[N];
bool cmpp(int a, int b) { return abs(ar[a] - ar[m]) < abs(ar[b] - ar[m]); }
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
      cin >> ar[i];
      pos[ar[i]] = i;
    }
    int md = pos[m];
    int ct = 0;
    br[ct]++;
    for (int i = md + 1; i <= n; i++) {
      if (ar[i] > m)
        ct++;
      else
        ct--;
      br[ct]++;
    }
    ans = br[0] + br[1];
    ct = 0;
    for (int i = md - 1; i > 0; i--) {
      if (ar[i] < m)
        ct++;
      else
        ct--;
      ans += br[ct] + br[ct + 1];
    }
    cout << ans << endl;
  }
  return 0;
}

```