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;
long long na, nb, k, a[100010], b[100010], i, j;
long long big[100010], sml[100010];
void I(long long *dat, long long x, long long d, long long v) {
  while (x > 0 && x <= nb) dat[x] += v, x += d * (x & -x);
}
long long Q(long long *dat, long long x, long long d) {
  long long ret = 0;
  while (x > 0 && x <= nb) ret += dat[x], x += d * (x & -x);
  return ret;
}
int main() {
  cin >> na >> k;
  for (i = 0; i < na; i++) cin >> a[i], b[i] = a[i];
  nb = unique(b, b + na) - b;
  sort(b, b + nb);
  for (i = 0; i < na; i++) a[i] = lower_bound(b, b + nb, a[i]) - b + 1;
  long long cnt = 0;
  for (i = na - 1; i >= 0; i--) {
    cnt += Q(sml, a[i] - 1, -1);
    I(sml, a[i], 1, 1);
  }
  I(sml, a[0], 1, -1);
  I(big, a[0], -1, 1);
  long long ans = 0;
  i = 0, j = 1;
  while (i < na && j < na) {
    while (i == j || cnt > k && j < na) {
      cnt -= Q(big, a[j] + 1, 1) + Q(sml, a[j] - 1, -1);
      I(sml, a[j++], 1, -1);
    }
    ans += na - j;
    cnt += Q(big, a[++i] + 1, 1) + Q(sml, a[i] - 1, -1);
    I(big, a[i], -1, 1);
  }
  cout << ans << endl;
}
