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;
using ll = long long;
template <class T>
using V = vector<T>;
template <class T, class U>
using P = pair<T, U>;
using vll = V<ll>;
using vvll = V<vll>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
const ll MOD = 1000000007;
const ll HIGHINF = (ll)1e18;
const int INF = 1e9;
int n, m;
V<int> a;
template <class T>
class SegmentTree {
  int n;
  vector<T> data;
  T def;
  function<T(T, T)> operation;
  function<T(T, T)> update;
  T _query(int a, int b, int k, int l, int r) {
    if (r <= a || b <= l) return def;
    if (a <= l && r <= b)
      return data[k];
    else {
      T c1 = _query(a, b, 2 * k + 1, l, (l + r) >> 1);
      T c2 = _query(a, b, 2 * k + 2, (l + r) >> 1, r);
      return operation(c1, c2);
    }
  }

 public:
  SegmentTree(size_t _n, T _def, function<T(T, T)> _operation,
              function<T(T, T)> _update)
      : def(_def), operation(_operation), update(_update) {
    n = 1;
    while (n < _n) n <<= 1;
    data = vector<T>(2 * n - 1, def);
  }
  void change(int i, T x) {
    i += n - 1;
    data[i] = update(data[i], x);
    while (i > 0) {
      i = (i - 1) >> 1;
      data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
    }
  }
  T query(int a, int b) { return _query(a, b, 0, 0, n); }
  T operator[](int i) { return data[i + n - 1]; }
};
ll count_med(int med) {
  V<int> pm1(n);
  for (int i = 0; i < n; i++) pm1[i] = (a[i] < med ? -1 : 1);
  ll ret = 0;
  int tmp = n;
  SegmentTree<int> seg(
      2 * n + 1, 0, [&](int i, int j) { return i + j; },
      [&](int i, int j) { return i + j; });
  for (int i = 0; i < n; i++) {
    seg.change(tmp, 1);
    tmp += pm1[i];
    ret += seg.query(0, tmp);
  }
  return ret;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cin >> n >> m;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  cout << count_med(m) - count_med(m + 1) << '\n';
  return 0;
}
