The Little Elephant has array a, consisting of n positive integers, indexed from 1 to n. Let's denote the number with index i as ai.

The Little Elephant wants to count, how many pairs of integers l and r are there, such that 1 ≤ l < r ≤ n and sequence b = a1a2... alarar + 1... an has no more than k inversions. 

An inversion in sequence b is a pair of elements of the sequence b, that change their relative order after a stable sorting of the sequence. In other words, an inversion is a pair of integers i and j, such that 1 ≤ i < j ≤ |b| and bi > bj, where |b| is the length of sequence b, and bj is its j-th element.

Help the Little Elephant and count the number of the described pairs.

Input

The first line contains two integers n and k (2 ≤ n ≤ 105, 0 ≤ k ≤ 1018) — the size of array a and the maximum allowed number of inversions respectively. The next line contains n positive integers, separated by single spaces, a1, a2, ..., an (1 ≤ ai ≤ 109) — elements of array a.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64d specifier.

Output

In a single line print a single number — the answer to the problem.

Examples

Input

3 1
1 3 2


Output

3


Input

5 2
1 3 2 1 7


Output

6
#include <bits/stdc++.h>
using namespace std;
struct BIT {
  int n;
  vector<int> bit;
  BIT(const int n) : n(n), bit(n + 1) {}
  inline int read(int i) {
    int sum = 0;
    while (i > 0) {
      sum += bit[i];
      i -= i & -i;
    }
    return sum;
  }
  inline int read_suffix(const int i) { return read(n) - read(i); }
  inline void update(int i, const int del = 1) {
    while (i <= n) {
      bit[i] += del;
      i += i & -i;
    }
  }
};
inline int compress(vector<int>& v) {
  map<int, int> d;
  int n = 0;
  for (const int e : v) d[e] = 1;
  for (auto& [a, b] : d) b = ++n;
  for (int& e : v) e = d[e];
  return n;
}
int main() {
  int n;
  long long k;
  cin >> n >> k;
  vector<int> v(n);
  for (int& e : v) {
    cin >> e;
  }
  const int net = compress(v);
  BIT left(net), right(net);
  long long res = 0, cur = 0;
  for (const int e : v) {
    cur += right.read_suffix(e);
    right.update(e, 1);
  }
  for (int i = 0, j = 0; j < n; i++) {
    cur += left.read_suffix(v[i]) + right.read(v[i] - 1);
    left.update(v[i], 1);
    while (j < n and (i == j or cur > k)) {
      cur -= right.read(v[j] - 1) + left.read_suffix(v[j]);
      right.update(v[j], -1);
      j++;
    }
    res += n - j;
  }
  cout << res << '\n';
}
