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 N = 300000;
long long k;
int n;
int t[N][2];
int sum(int r, int z) {
  int result = 0;
  for (; r >= 0; r = (r & (r + 1)) - 1) result += t[r][z];
  return result;
}
void inc(int i, int delta, int z) {
  for (; i < n; i = (i | (i + 1))) t[i][z] += delta;
}
pair<int, int> a[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    a[i] = make_pair(x, i);
  }
  sort(a, a + n);
  int p = -1;
  int s = -1;
  for (int i = 0; i < n; i++) {
    if (a[i].first == p)
      a[i].first = s;
    else {
      p = a[i].first;
      a[i].first = ++s;
    }
    swap(a[i].first, a[i].second);
  }
  sort(a, a + n);
  int l = 0;
  int r = n - 1;
  inc(a[l].second, 1, 1);
  long long q = 0;
  while (r > l) {
    if (q + sum(n - 1, 1) - sum(a[r].second, 1) + sum(a[r].second - 1, 0) <=
        k) {
      q += sum(n - 1, 1) - sum(a[r].second, 1) + sum(a[r].second - 1, 0);
      inc(a[r].second, 1, 0);
      r--;
    } else
      break;
  }
  long long ans = n - 1 - r;
  while (l < n - 1) {
    l++;
    inc(a[l].second, 1, 1);
    q += sum(a[l].second - 1, 0) + sum(n - 1, 1) - sum(a[l].second, 1);
    while (r < l || q > k) {
      if (r == n - 1) break;
      r++;
      q -= sum(n - 1, 1) - sum(a[r].second, 1) + sum(a[r].second - 1, 0);
      inc(a[r].second, -1, 0);
      if (r == n - 1) break;
    }
    ans += n - 1 - r;
  }
  cout << ans;
}
