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;
void toFile(string name) {
  freopen((name + ".inp").c_str(), "r", stdin);
  freopen((name + ".out").c_str(), "w", stdout);
}
long long ceil(long long a, long long x) {
  return (a % x == 0 ? a / x : a / x + 1);
}
const double pi = 3.141592653589;
const long long mod = 1e9 + 7;
const long long inf = 0x3f3f3f3f3f3f3f3f;
const double eps = 1e-6;
const long long mxN = 1e6 + 5;
long long n, m;
long long a[mxN];
long long bit[mxN];
long long bit2[mxN];
void update(long long i, long long b) {
  for (; i < 1e6; i = (i | (i + 1))) {
    bit[i] += b;
  }
}
long long sum(long long i) {
  long long res = 0;
  for (; ~i; i = (i & (i + 1)) - 1) {
    res += bit[i];
  }
  return res;
}
void update2(long long i, long long b) {
  for (; i < 1e6; i = (i | (i + 1))) {
    bit2[i] += b;
  }
}
long long sum2(long long i) {
  long long res = 0;
  for (; ~i; i = (i & (i + 1)) - 1) {
    res += bit2[i];
  }
  return res;
}
long long findans(long long m) {
  long long res = 0;
  memset(bit, 0, sizeof(bit));
  memset(bit2, 0, sizeof(bit2));
  pair<long long, long long> p = {0, 0};
  for (long long i = 0; i < n; i++) {
    if (a[i] <= m)
      p.first++;
    else
      p.second++;
    long long s = 0;
    if (i & 1) {
      if (p.first - p.second >= 0) res++;
      res += sum(p.first - p.second + 3e5);
      res += sum2(p.first - p.second - 1 + 3e5);
      update(p.first - p.second + 3e5, 1);
    } else {
      if (p.first - p.second >= 1) res++, s++;
      res += sum2(p.first - p.second + 3e5);
      res += sum(p.first - p.second - 1 + 3e5);
      update2(p.first - p.second + 3e5, 1);
    }
  }
  return res;
}
void solve() {
  cin >> n >> m;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << findans(m) - findans(m - 1);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  solve();
  return 0;
}
