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>
#pragma GCC optimize("O3")
using namespace std;
const long long mod7 = 1000000007LL;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
inline long long add(long long _x, long long _y, long long _mod = mod7) {
  _x += _y;
  return _x >= _mod ? _x - _mod : _x;
}
inline long long sub(long long _x, long long _y, long long _mod = mod7) {
  _x -= _y;
  return _x < 0 ? _x + _mod : _x;
}
inline long long mul(long long _x, long long _y, long long _mod = mod7) {
  _x *= _y;
  return _x >= _mod ? _x % _mod : _x;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1ll;
  long long _ret = mypow(mul(_a, _a, _mod), _x >> 1, _mod);
  if (_x & 1) _ret = mul(_ret, _a, _mod);
  return _ret;
}
long long mymul(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 0ll;
  long long _ret = mymul(add(_a, _a, _mod), _x >> 1, _mod);
  if (_x & 1) _ret = add(_ret, _a, _mod);
  return _ret;
}
void sleep(double sec = 1021) {
  clock_t s = clock();
  while (clock() - s < CLOCKS_PER_SEC * sec)
    ;
}
int __ = 1, _cs;
const int mxN = 2e5;
int n, m, a[mxN], ps1, ps2;
vector<int> pts[2 * mxN + 1], fts[2 * mxN + 1];
long long ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; ++i) {
    for (int j = n + ps2 - ps1; j <= 2 * n; j += j & -j)
      pts[j].push_back(1 + ps2 + ps1);
    cin >> a[i];
    if (a[i] > m)
      ++ps1;
    else if (a[i] < m)
      --ps1;
    else
      ++ps2;
  }
  for (int i = 1; i <= 2 * n; ++i) {
    sort(pts[i].begin(), pts[i].end());
    fts[i] = vector<int>(pts[i].size() + 1);
  }
  ps1 = 0;
  ps2 = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = n + ps2 - ps1; j <= 2 * n; j += j & -j)
      for (int k = lower_bound(pts[j].begin(), pts[j].end(), 1 + ps2 + ps1) -
                   pts[j].begin() + 1;
           k < fts[j].size(); k += k & -k)
        ++fts[j][k];
    if (a[i] > m)
      ++ps1;
    else if (a[i] < m)
      --ps1;
    else
      ++ps2;
    for (int j = n + ps2 - ps1; j; j -= j & -j)
      for (int k = upper_bound(pts[j].begin(), pts[j].end(), ps2 + ps1) -
                   pts[j].begin();
           k; k -= k & -k)
        ans += fts[j][k];
  }
  cout << ans;
}
