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).
#include <bits/stdc++.h>
using namespace std;
const long long N = 200000 + 5;
struct BIT {
  vector<long long> bit;
  long long n;
  void init(long long _n) {
    bit.clear();
    bit.resize(_n + 10, 0);
    n = _n;
  }
  void update(long long i, long long x) {
    while (i <= n) {
      bit[i] += x;
      i += i & -i;
    }
  }
  long long query(long long i) {
    long long res = 0;
    while (i > 0) {
      res += bit[i];
      i -= i & -i;
    }
    return res;
  }
} bit;
vector<long long> a;
vector<long long> b;
signed main() {
  long long n, m;
  cin >> n >> m;
  a.resize(n + 10);
  for (long long i = 1; i < n + 1; i++) cin >> a[i];
  b = a;
  bit.init(2 * n + 10);
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 1; i < n + 1; i++) {
    b[i] = (b[i] >= m ? 1 : -1) + b[i - 1];
    bit.update(b[i] + n + 1, 1);
    cnt1 += bit.query(n + b[i]);
    if (b[i] > 0) cnt1++;
  }
  b = a;
  bit.init(2 * n + 10);
  for (long long i = 1; i < n + 1; i++) {
    b[i] = (b[i] > m ? 1 : -1) + b[i - 1];
    bit.update(b[i] + n + 1, 1);
    cnt2 += bit.query(n + b[i]);
    if (b[i] > 0) cnt2++;
  }
  cout << cnt1 - cnt2;
}
