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;
#pragma warning(disable : 4996)
const int IT_MAX = 1 << 18;
const int MAXN = 2e5 + 5;
const int mod = 1e9 + 7;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
long long powmod(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
struct bit {
  int tree[MAXN << 1], lim = (MAXN << 1) - 1;
  void init() { memset(tree, 0, sizeof(tree)); }
  void add(int x) {
    while (x <= lim) tree[x]++, x += x & -x;
  }
  int sum(int x) {
    int ret = 0;
    while (x) ret += tree[x], x -= x & -x;
    return ret;
  }
} bit;
int n, m, b[MAXN], a[MAXN];
long long f(int x) {
  long long ret = 0;
  bit.init();
  for (int i = 1; i < n + 1; i++) {
    b[i] = ((a[i] <= x) ? 1 : -1);
    b[i] += b[i - 1];
  }
  bit.add(n + 1);
  for (int i = 1; i < n + 1; i++) {
    ret += bit.sum(b[i] + n + 1);
    bit.add(n + 1 + b[i]);
  }
  return ret;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i < n + 1; i++) scanf("%d", a + i);
  printf("%lld\n", f(m) - f(m - 1));
}
