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 tree[2][120000], n, l, r, p[120000], a[120000], i;
long long ans, k, v;
int der(int x) { return lower_bound(p + 1, p + n + 1, x) - p; }
int calc(int o, int x) {
  if (!o) x = n - x + 1;
  int res = 0;
  for (int i = x; i; i -= i & (-i)) res += tree[o][i];
  return res;
}
void add(int o, int x, int flag) {
  if (!o) x = n - x + 1;
  for (int i = x; i <= n; i += i & (-i)) tree[o][i] += flag;
}
int main() {
  scanf("%d%I64d", &n, &k);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), p[i] = a[i];
  sort(p + 1, p + n + 1);
  for (i = n; i >= 1; --i) {
    a[i] = der(a[i]);
    v += calc(1, a[i] - 1);
    add(1, a[i], 1);
  }
  r = 1;
  for (l = 1; l <= n; ++l) {
    v += calc(0, a[l] + 1) + calc(1, a[l] - 1);
    add(0, a[l], 1);
    while ((r <= l || v > k) && r <= n) {
      v -= calc(0, a[r] + 1) + calc(1, a[r] - 1);
      add(1, a[r], -1);
      r++;
    }
    ans += n - r + 1;
  }
  printf("%I64d\n", ans);
}
