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;
int64_t mod = 1e9 + 7;
int64_t mod1 = 1e9 + 5;
int64_t power(int64_t a, int64_t b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power(((((a) % mod) * ((a) % mod)) % mod), b / 2) % mod;
  else
    return (((a) % mod) *
            (power(((((a) % mod) * ((a) % mod)) % mod), b / 2) % mod)) %
           mod;
}
inline int64_t inverse(int64_t a, int64_t md) {
  a %= md;
  if (a < 0) a += md;
  int64_t b = md, u = 0, v = 1;
  while (a) {
    int64_t t = b / a;
    b -= t * a;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  assert(b == 1);
  if (u < 0) u += md;
  return u;
}
const int64_t ce = 1e9 + 7;
int64_t fast_mod(int64_t input) { return input < ce ? input : input % ce; }
int64_t a[200001];
int64_t pre[200001];
int64_t n;
vector<int64_t> s(500001, 0);
void update(int64_t pos, int64_t dif) {
  for (; pos < (s).size(); pos |= pos + 1) s[pos] = (s[pos] + dif);
}
int64_t query(int64_t pos) {
  int64_t res = 0;
  for (; pos > 0; pos &= pos - 1) res = (res + s[pos - 1]);
  return res;
}
int64_t find(int64_t k) {
  pre[0] = 0;
  for (int64_t i = 0; i <= 500000; i++) s[i] = 0;
  for (int64_t i = 1; i <= n; i++) {
    pre[i] = pre[i - 1];
    if (a[i] >= k)
      pre[i]++;
    else
      pre[i]--;
    update(pre[i] + 200000, 1);
  }
  update(pre[0] + 200000, 1);
  int64_t ans = 0;
  for (int64_t i = 1; i <= n; i++) {
    update(pre[i - 1] + 200000, -1);
    ans += query(500000) - query(pre[i - 1] + 1 + 200000);
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int64_t t;
  t = 1;
  while (t--) {
    int64_t k;
    cin >> n >> k;
    for (int64_t i = 1; i <= n; i++) cin >> a[i];
    cout << find(k) - find(k + 1) << '\n';
  }
}
