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 = 2e5 + 20;
int a[maxn], fen[maxn][2], b[maxn];
void add(int pos, int type, int val) {
  for (pos++; pos < maxn; pos += ((pos) & (-pos))) fen[pos][type] += val;
}
int get(int pos, int type) {
  int res = 0;
  for (pos++; pos > 0; pos -= ((pos) & (-pos))) res += fen[pos][type];
  return res;
}
int main() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++) a[i] = lower_bound(b, b + n, a[i]) - b;
  long long ans = 0, inv = 0, R = 1;
  for (int i = n - 1; i >= 1; i--) {
    inv += get(a[i] - 1, 1);
    add(a[i], 1, 1);
  }
  for (int i = 0; i < n; i++) {
    inv += get(a[i] - 1, 1);
    inv += get(maxn - 2, 0) - get(a[i], 0);
    add(a[i], 0, 1);
    while (R <= i || (R < n && inv > k)) {
      inv -= get(a[R] - 1, 1);
      inv -= get(maxn - 2, 0) - get(a[R], 0);
      add(a[R], 1, -1);
      R++;
    }
    ans += n - R;
  }
  cout << ans << endl;
}
