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 INF = 1e9;
const long long INFF = 1e18;
const long long MAXN = 510;
const long long MOD = 1e9 + 7;
const double PI = acos(-1.0);
const double INFD = 1E9;
const double EPS = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long getTC() {
  if (0) {
    long long t;
    cin >> t;
    return t;
  } else
    return 1;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> tab(n);
  for (auto &it : tab) cin >> it;
  const auto cnt = [&](long long x) -> long long {
    vector<long long> frek(2 * n + 1, 0);
    long long cur = n;
    long long ret = 0;
    frek[n] = 1;
    long long delta = 0;
    for (long long i = 0; i < n; i++) {
      if (tab[i] < x) {
        --cur;
        delta -= frek[cur];
      } else {
        delta += frek[cur];
        ++cur;
      }
      ret += delta;
      frek[cur]++;
    }
    return ret;
  };
  long long ans = cnt(m) - cnt(m + 1);
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = getTC();
  while (t--) {
    solve();
  }
  return 0;
}
