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;
int i, r, n, x, a[100100], s[100100], t[100100];
pair<int, int> y[100100];
long long m, ans, cur, fl[100100], fr[100100];
void modify(int *s, int i, int v) {
  for (; i <= n; i = (i << 1) - (i & (i - 1))) s[i] += v;
}
int cnt(int *s, int i) {
  int r = 0;
  for (; i > 0; i &= i - 1) r += s[i];
  return r;
}
void remove(int n) {
  modify(t, a[r], -1);
  cur -= n - cnt(s, a[r]);
  r++;
}
int main() {
  scanf("%d%I64d", &n, &m);
  for (i = 1; i <= n; i++) {
    scanf("%d", &x);
    y[i - 1] = make_pair(x, i);
  }
  sort(y, y + n);
  for (i = 0; i < n; i++) a[y[i].second] = i + 1;
  for (i = 1; i <= n; i++) {
    fl[i] = fl[i - 1] + i - 1 - cnt(s, a[i]);
    modify(s, a[i], 1);
  }
  for (i = 1; i <= n; i++) s[i] = 0;
  for (i = n; i >= 1; i--) {
    fr[i] = fr[i + 1] + cnt(s, a[i]);
    modify(s, a[i], 1);
  }
  for (i = 1; i <= n; i++) s[i] = 0;
  for (i = n; i >= 2; i--) modify(t, a[i], 1);
  for (i = 1, r = 2; i < n; i++) {
    if (r == i) remove(i - 1);
    modify(s, a[i], 1);
    cur += cnt(t, a[i]);
    while (r <= n) {
      if (fl[i] + fr[r] + cur <= m) break;
      remove(i);
    }
    ans += n - r + 1;
  }
  printf("%I64d\n", ans);
  return 0;
}
