#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 42;
const int N = 5e5 + 42;
int one[N];
int two[N];
void add(int *fen, int i, int v) {
  for (i += N / 2; i < N; i += i & -i) fen[i] += v;
}
int sum(int *fen, int i) {
  int v = 0;
  for (i += N / 2; i > 0; i -= i & -i) v += fen[i];
  return v;
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  long long ans = 0;
  int p = 0, q = 0;
  add(one, p, 1);
  add(two, q, 1);
  for (int &x : a) {
    cin >> x;
    if (x < m)
      p--, q--;
    else if (x > m)
      p++, q++;
    else
      p++, q--;
    add(one, p, 1);
    add(two, q, 1);
    ans += sum(one, p - 1) - sum(two, q - 1);
  }
  kek(ans);
}

### Prompt

Please provide a CPP coded solution to the problem described below:

You are given an integer sequence a_1, a_2, ..., a_n.

Find the number of pairs of indices (l, r) (1 ≤ l ≤ r ≤ n) such that the value of median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

The median of a sequence is the value of an 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 median of a_l, a_{l+1}, ..., a_r is exactly the given number m.

Input

The first line contains integers n and m (1 ≤ n,m ≤ 2⋅10^5) — the length of the given sequence and the required value of the median.

The second line contains an integer sequence a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2⋅10^5).

Output

Print the required number.

Examples

Input

5 4
1 4 5 60 4


Output

8


Input

3 1
1 1 1


Output

6


Input

15 2
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3


Output

97

Note

In the first example, the suitable pairs of indices are: (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 5), (4, 5) and (5, 5).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline bool uax(T &x, T y) {
  return (y > x) ? x = y, true : false;
}
template <typename T>
inline bool uin(T &x, T y) {
  return (y < x) ? x = y, true : false;
}
template <typename T>
void kek(T ans) {
  cout << ans << endl;
  exit(0);
}
const int MOD = (int)1e9 + 7;
const long long INF = (long long)1e18 + 42;
const int N = 5e5 + 42;
int one[N];
int two[N];
void add(int *fen, int i, int v) {
  for (i += N / 2; i < N; i += i & -i) fen[i] += v;
}
int sum(int *fen, int i) {
  int v = 0;
  for (i += N / 2; i > 0; i -= i & -i) v += fen[i];
  return v;
}
int main() {
  cin.tie(nullptr)->sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  long long ans = 0;
  int p = 0, q = 0;
  add(one, p, 1);
  add(two, q, 1);
  for (int &x : a) {
    cin >> x;
    if (x < m)
      p--, q--;
    else if (x > m)
      p++, q++;
    else
      p++, q--;
    add(one, p, 1);
    add(two, q, 1);
    ans += sum(one, p - 1) - sum(two, q - 1);
  }
  kek(ans);
}

```