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 n, m;
long long k;
int a[100020], b[100020];
int c1[100020], c2[100020];
void R(int *a, int x, int y) {
  for (; x <= m; x += x & -x) a[x] += y;
}
int G(int *a, int x) {
  int _ = 0;
  for (; x; x -= x & -x) _ += a[x];
  return _;
}
long long s, z;
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), b[m++] = a[i];
  sort(b, b + m);
  m = unique(b, b + m) - b;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b, b + m, a[i]) - b + 1;
  for (int i = 1; i <= n; i++) {
    R(c1, a[i], 1);
    s += i - G(c1, a[i]);
  }
  if (s <= k) return cout << (long long)n * (n - 1) / 2, 0;
  R(c1, a[n], -1);
  R(c2, a[n], 1);
  cerr << s << endl;
  for (int r = n, l = n - 1;;) {
    while (l > 0 && s > k) {
      s -= l - G(c1, a[l]);
      s -= G(c2, a[l] - 1);
      R(c1, a[l], -1);
      l--;
    }
    if (l == 0) break;
    z += l;
    r--;
    s += l - G(c1, a[r]);
    s += G(c2, a[r] - 1);
    R(c2, a[r], 1);
  }
  printf("%I64d", z);
  return 0;
}
