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;
long long mod = 1e9 + 7;
long long min(long long a, long long b) { return (a < b) ? a : b; }
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long fp(long long a, long long b) {
  if (b == 0) return 1;
  long long x = fp(a, b / 2);
  x = (x * x) % mod;
  if (b & 1) x = (x * a) % mod;
  return x;
}
auto random_add = [] {
  char *p = new char;
  delete p;
  return uint64_t(p);
};
const uint64_t SEED =
    chrono::steady_clock::now().time_since_epoch().count() * (random_add() | 1);
mt19937_64 rng(SEED);
const long long N = 2e5 + 5;
long long solve1(long long m, vector<long long> v) {
  map<long long, long long> mp;
  mp[0] = 1;
  long long sm = 0, ans = 0, x = 0;
  long long n = v.size();
  for (long long i = 0; i < n; i++) {
    if (v[i] >= m) {
      x++;
      sm += mp[x - 1];
    } else {
      sm -= mp[x - 1];
      x--;
    }
    ans += sm;
    mp[x]++;
  }
  return ans;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  long long ans = solve1(m, a) - solve1(m + 1, a);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  srand(time(0));
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
}
