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;
template <typename T>
T gcd(T m, T n) {
  while (n) {
    T t = m % n;
    m = n;
    n = t;
  };
  return m;
}
template <typename T>
T exgcd(T a, T b, T& sa, T& ta) {
  T q, r, sb = 0, tb = 1, sc, tc;
  sa = 1, ta = 0;
  if (b) do
      q = a / b, r = a - q * b, a = b, b = r, sc = sa - q * sb, sa = sb,
      sb = sc, tc = ta - q * tb, ta = tb, tb = tc;
    while (b);
  return a;
}
template <typename T>
T mul_inv(T a, T b) {
  T t1 = a, t2 = b, t3;
  T v1 = 1, v2 = 0, v3;
  T x;
  while (t2 != 1)
    x = t1 / t2, t3 = t1 - x * t2, v3 = v1 - x * v2, t1 = t2, t2 = t3, v1 = v2,
    v2 = v3;
  return (v2 + b) % b;
}
template <typename T>
T powmod(T a, T b, T MOD) {
  if (b < 0) return 0;
  T rv = 1;
  while (b) (b % 2) && (rv = (rv * a) % MOD), a = a * a % MOD, b /= 2;
  return rv;
}
template <long long n>
struct BitSize {
  enum { Size = BitSize<n / 2>::Size + 1 };
};
template <>
struct BitSize<0> {
  enum { Size = 1 };
};
template <>
struct BitSize<1> {
  enum { Size = 1 };
};
template <typename TH>
void _dbg(const char* sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char* sdbg, TH h, TA... t) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, t...);
}
const int maxn = 2e5 + 7;
const int maxm = 2e5 + 7;
int n, m;
int a[maxn];
int aa[maxn];
int s[maxn];
template <typename T, int MAXM, int OFF>
class BIT {
 public:
  enum { MX = MAXM - 1 };

 private:
  int m[MAXM + OFF];
  inline T acc(T& r, const T& x) const { return r + x; }

 public:
  void reset() { memset(m, 0, sizeof(m)); }
  inline void update(int x, T v) {
    x += OFF;
    while (x < MAXM + OFF) m[x] = acc(m[x], v), x += x & -x;
  }
  inline T query(int x) {
    x += OFF;
    T r = 0;
    while (x > 0) r = acc(r, m[x]), x -= x & -x;
    return r;
  }
};
BIT<int, maxn + 7, maxn + 7> b;
int main(int argc, char* argv[]) {
  std::cin.sync_with_stdio(false);
  std::cin.tie(nullptr);
  cin >> n >> m;
  s[0] = 0;
  b.reset();
  b.update(0, 1);
  long long r = 0;
  for (int i = 1; i <= n; i++) {
    cin >> aa[i];
    if (aa[i] >= m)
      a[i] = 1;
    else
      a[i] = -1;
    s[i] = s[i - 1] + a[i];
    r += b.query(s[i] - 1);
    b.update(s[i], 1);
  }
  _dbg("r", r);
  b.reset();
  b.update(0, 1);
  for (int i = 1; i <= n; i++) {
    if (aa[i] <= m)
      a[i] = 1;
    else
      a[i] = -1;
    s[i] = s[i - 1] + a[i];
    r += b.query(s[i]);
    b.update(s[i], 1);
  }
  cout << (r - (long long)n * (n + 1) / 2) << endl;
  return 0;
}
