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 long long mod = 1e9 + 7;
int n;
const int N = 1e5 + 5;
long long c[N][2];
int lb(int x) { return x & (-x); }
void add(int i, long long x, int fg) {
  while (i <= n) c[i][fg] += x, i += lb(i);
}
long long sum(int i, int fg) {
  if (i < 0) return 0;
  long long res = 0;
  while (i > 0) {
    res += c[i][fg];
    i -= lb(i);
  }
  return res;
}
int b[N], a[N];
int main(int argc, char** argv) {
  long long k;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + n + 1);
  unique(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;
  long long nx = 0, ans = 0;
  for (int i = 1; i <= n; i++) {
    nx += sum(n, 1) - sum(a[i], 1);
    add(a[i], 1, 1);
  }
  int j = 1;
  for (int i = 2; i <= n; i++) {
    add(a[i - 1], -1, 1);
    nx -= sum(n, 0) - sum(a[i - 1], 0) + sum(a[i - 1] - 1, 1);
    while (j < i && nx <= k) {
      add(a[j], 1, 0);
      nx += sum(n, 0) - sum(a[j], 0) + sum(a[j] - 1, 1);
      j++;
    }
    if (nx > k)
      ans += max(0, j - 2);
    else
      ans += j - 1;
  }
  cout << ans;
  return 0;
}
