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;
const int maxn = 100000 + 100;
int n, nb;
long long k;
int a[maxn], b[maxn];
int fl[maxn], fr[maxn];
void add(int *f, int p, int x) {
  for (p++; p <= nb; p += p & -p) f[p] += x;
}
int get(int *f, int p) {
  int r = 0;
  for (p++; p; p -= p & -p) r += f[p];
  return r;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0, _n = (int)(n); i < _n; i++) cin >> a[i];
  copy(a, a + n, b);
  sort(b, b + n);
  nb = unique(b, b + n) - b;
  for (int i = 0, _n = (int)(n); i < _n; i++)
    a[i] = lower_bound(b, b + nb, a[i]) - b;
  long long ans = 0;
  long long cur = 0;
  int r;
  for (r = n; --r;) cur += get(fr, a[r] - 1), add(fr, a[r], 1);
  r++;
  for (int l = 0, _n = (int)(n); l < _n; l++) {
    if (l == r) {
      cur -= get(fr, a[r] - 1) + l - get(fl, a[r]);
      add(fr, a[r], -1);
      r++;
    }
    cur += l - get(fl, a[l]) + get(fr, a[l] - 1);
    add(fl, a[l], 1);
    while (r < n && cur > k) {
      cur -= get(fr, a[r] - 1) + (l + 1) - get(fl, a[r]);
      add(fr, a[r], -1);
      r++;
    }
    ans += n - r;
  }
  cout << ans << endl;
  {
    int _;
    cin >> _;
  }
  return 0;
}
