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, i, l, r, a[100010], aib1[100010], aib2[100010];
long long k, val, sol;
pair<int, int> b[100010];
void upd(int poz, int val) {
  for (; poz <= n; poz += poz & -poz) aib1[poz] += val;
}
long long getsum(int poz) {
  long long ret = 0;
  for (; poz; poz -= poz & -poz) ret += aib1[poz];
  return ret;
}
void upd2(int poz, int val) {
  poz = n - poz + 1;
  for (; poz <= n; poz += poz & -poz) aib2[poz] += val;
}
long long getsum2(int poz) {
  poz = n - poz + 1;
  long long ret = 0;
  for (; poz; poz -= poz & -poz) ret += aib2[poz];
  return ret;
}
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    b[i].first = a[i];
    b[i].second = i;
  }
  sort(b + 1, b + n + 1);
  for (i = 1; i <= n; i++)
    if (b[i].first == b[i - 1].first)
      a[b[i].second] = a[b[i - 1].second];
    else
      a[b[i].second] = a[b[i - 1].second] + 1;
  for (i = n; i >= 1; i--) {
    val += getsum(a[i] - 1);
    upd(a[i], 1);
  }
  upd(a[1], -1);
  upd2(a[1], 1);
  for (r = 2, l = 1; l < n; l++) {
    while ((val > k && r < n) || (r == l)) {
      val -= getsum(a[r] - 1);
      val -= getsum2(a[r] + 1);
      upd(a[r], -1);
      r++;
    }
    if (val <= k)
      sol += n - r + 1;
    else
      break;
    val += getsum(a[l + 1] - 1);
    val += getsum2(a[l + 1] + 1);
    upd2(a[l + 1], 1);
  }
  cout << sol;
  return 0;
}
