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 int inf = 0x3f3f3f3f;
const long long mod = 1000000007;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const int N = 300100;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
struct segtree {
  int N;
  long long first[1600200];
  long long lazy[1600200];
  void push(int node, int lo, int hi) {
    if (lazy[node] == 0) return;
    first[node] += lazy[node];
    if (lo != hi) {
      lazy[2 * node + 1] += lazy[node];
      lazy[2 * node + 2] += lazy[node];
    }
    lazy[node] = 0;
  }
  void update(int s, int e, long long x, int lo = 0, int hi = -1,
              int node = 0) {
    if (hi == -1) hi = N - 1;
    push(node, lo, hi);
    if (hi < s || lo > e) return;
    if (lo >= s && hi <= e) {
      lazy[node] = x;
      push(node, lo, hi);
      return;
    }
    int mid = (lo + hi) / 2;
    update(s, e, x, lo, mid, 2 * node + 1);
    update(s, e, x, mid + 1, hi, 2 * node + 2);
    first[node] = max(first[2 * node + 1], first[2 * node + 2]);
  }
  long long query(int s, int e, int lo = 0, int hi = -1, int node = 0) {
    if (hi == -1) hi = N - 1;
    push(node, lo, hi);
    if (hi < s || lo > e) return 0;
    if (lo >= s && hi <= e) return first[node];
    int mid = (lo + hi) / 2;
    return max(query(s, e, lo, mid, 2 * node + 1),
               query(s, e, mid + 1, hi, 2 * node + 2));
  }
};
int n, k;
int a[N];
int c[N];
int pref[N];
segtree seg;
int MX = 200000;
long long sol(int x) {
  for (int i = 0; i < 1600200; i++) seg.first[i] = 0, seg.lazy[i] = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] < x)
      c[i] = -1;
    else
      c[i] = 1;
  }
  long long ans = 0;
  seg.update(MX, MX + MX, 1);
  for (int i = 0; i < n; i++) {
    pref[i + 1] = pref[i] + c[i];
    int aux = MX + pref[i + 1] - 1;
    ans += seg.query(aux, aux);
    seg.update(MX + pref[i + 1], MX + MX, 1);
  }
  return ans;
}
void solve() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  seg.N = 400010;
  long long ans = sol(k) - sol(k + 1);
  cout << ans << "\n";
}
int main() {
  cin.tie(0)->sync_with_stdio(0);
  ;
  int t_ = 1;
  while (t_--) solve();
  return 0;
}
