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;
using VS = vector<string>;
using LL = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using VL = vector<LL>;
using VVL = vector<VL>;
const int INF = 1e9;
const LL LINF = 1e16;
const LL MOD = 1000000007;
const double PI = acos(-1.0);
int DX[8] = {0, 0, 1, -1, 1, 1, -1, -1};
int DY[8] = {1, -1, 0, 0, 1, -1, 1, -1};
template <typename type>
struct BIT {
  int N;
  int nn;
  vector<type> data;
  BIT(int n) {
    N = n + 1;
    data = vector<type>(n + 1, 0);
    nn = 1;
    while (nn * 2 <= N) nn *= 2;
  }
  void add(int i, type w) {
    for (int x = i; x <= N; x += x & -x) {
      data[x] += w;
    }
  }
  type sum(int i) {
    type ret = 0;
    for (int x = i; x > 0; x -= x & -x) {
      ret += data[x];
    }
    return ret;
  }
  type sum(int l, int r) {
    if (l > r) return 0;
    return sum(r) - sum(l - 1);
  }
};
LL gre(const VI& a, int M) {
  int N = int((a).size());
  LL ret = 0;
  BIT<LL> bit(4e5 + 10);
  int cur = 2e5 + 5;
  for (int(i) = (0); (i) < (N); (i)++) {
    bit.add(cur, 1);
    if (a[i] < M) {
      cur--;
    } else {
      cur++;
    }
    ret += bit.sum(cur - 1);
  }
  return ret;
}
void solve2() {
  int N, M;
  cin >> N >> M;
  VI a(N);
  for (int(i) = (0); (i) < (N); (i)++) cin >> a[i];
  cout << gre(a, M) - gre(a, M + 1) << endl;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  solve2();
  return 0;
}
