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;
void update(int64_t fentree[], int64_t n, int64_t index, int64_t val) {
  while (index <= n) {
    fentree[index] += val;
    index += index & (-index);
  }
}
int64_t get_sum(int64_t fentree[], int64_t index) {
  int64_t sum = 0;
  while (index > 0) {
    sum += fentree[index];
    index -= index & (-index);
  }
  return sum;
}
void convert(int64_t a[], int64_t n) {
  int64_t temp[n], i;
  for (int64_t i = int64_t(0); i < int64_t(n); ++i) temp[i] = a[i];
  sort(temp, temp + n);
  for (int64_t i = int64_t(0); i < int64_t(n); ++i)
    a[i] = lower_bound(temp, temp + n, a[i]) - temp + 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int64_t z, n, m, t, k, i, j, l, d, h, r;
  cin >> n >> k;
  int64_t a[n];
  for (int64_t i = int64_t(0); i < int64_t(n); ++i) cin >> a[i];
  convert(a, n);
  int64_t fentree_right[n + 1], fentree_left[n + 1];
  for (int64_t i = int64_t(0); i < int64_t(n + 1); ++i) {
    fentree_left[i] = 0;
    fentree_right[i] = 0;
  }
  z = 0;
  int64_t ans = 0, cnt = 0;
  for (int64_t i = int64_t(n - 1); i >= int64_t(0); --i) {
    r = i;
    z = get_sum(fentree_right, a[i] - 1);
    if (z + cnt > k) {
      r++;
      break;
    }
    cnt += z;
    update(fentree_right, n, a[i], 1);
  }
  for (int64_t i = int64_t(0); i < int64_t(n); ++i) {
    cnt += get_sum(fentree_right, a[i] - 1) + i - get_sum(fentree_left, a[i]);
    update(fentree_left, n, a[i], 1);
    while ((cnt > k or r <= i) and r <= n - 1) {
      z = get_sum(fentree_right, a[r] - 1) + i + 1 -
          get_sum(fentree_left, a[r]);
      cnt -= z;
      update(fentree_right, n, a[r], -1);
      r++;
    }
    ans += n - r;
  }
  cout << ans << "\n";
  return 0;
}
