#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 2e5 + 5;
const int MAXM = 2e6 + 5;
const int MOD = 1e9 + 7;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long pw(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void inc(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void dec(long long &a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mul(int a, int b) {
  long long c = 1ll * a * b;
  return c - c / MOD * MOD;
}
int n, m, k, a[MAXN], s[MAXN], c[MAXN];
vector<int> ve;
int getid(int x) {
  return lower_bound((ve).begin(), (ve).end(), x) - ve.begin() + 1;
}
void upd(int x) {
  for (; x <= k; x += x & -x) c[x]++;
}
int qry(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret += c[x];
  return ret;
}
long long get(int x) {
  ve.clear();
  memset(c, 0, sizeof(c));
  s[0] = 0;
  ve.push_back(0);
  for (int i = 1; i < n + 1; i++)
    s[i] = s[i - 1] + (a[i] >= x ? 1 : -1), ve.push_back(s[i]);
  sort((ve).begin(), (ve).end());
  ve.erase(unique((ve).begin(), (ve).end()), ve.end());
  k = ((int)(ve).size());
  long long ret = 0;
  upd(getid(0));
  for (int i = 1; i < n + 1; i++) {
    int pos = getid(s[i]);
    ret += qry(pos - 1);
    upd(pos);
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  printf("%lld\n", get(m) - get(m + 1));
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 INF = 0x3f3f3f3f;
const double eps = 1e-8;
const double PI = acos(-1.0);
const int MAXN = 2e5 + 5;
const int MAXM = 2e6 + 5;
const int MOD = 1e9 + 7;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long pw(long long a, long long n, long long mod) {
  long long ret = 1;
  while (n) {
    if (n & 1) ret = ret * a % mod;
    a = a * a % mod;
    n >>= 1;
  }
  return ret;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
void inc(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
void dec(long long &a, long long b) {
  a -= b;
  if (a < 0) a += MOD;
}
int mul(int a, int b) {
  long long c = 1ll * a * b;
  return c - c / MOD * MOD;
}
int n, m, k, a[MAXN], s[MAXN], c[MAXN];
vector<int> ve;
int getid(int x) {
  return lower_bound((ve).begin(), (ve).end(), x) - ve.begin() + 1;
}
void upd(int x) {
  for (; x <= k; x += x & -x) c[x]++;
}
int qry(int x) {
  int ret = 0;
  for (; x; x -= x & -x) ret += c[x];
  return ret;
}
long long get(int x) {
  ve.clear();
  memset(c, 0, sizeof(c));
  s[0] = 0;
  ve.push_back(0);
  for (int i = 1; i < n + 1; i++)
    s[i] = s[i - 1] + (a[i] >= x ? 1 : -1), ve.push_back(s[i]);
  sort((ve).begin(), (ve).end());
  ve.erase(unique((ve).begin(), (ve).end()), ve.end());
  k = ((int)(ve).size());
  long long ret = 0;
  upd(getid(0));
  for (int i = 1; i < n + 1; i++) {
    int pos = getid(s[i]);
    ret += qry(pos - 1);
    upd(pos);
  }
  return ret;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i < n + 1; i++) scanf("%d", &a[i]);
  printf("%lld\n", get(m) - get(m + 1));
}

```