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 c[2][100010], n, cnt, a[100010], b[100010];
long long k, ans, sum;
int query(int tmp, int x) {
  int sum = 0;
  for (; x; x -= (x & (-x))) sum += c[tmp][x];
  return sum;
}
void add(int tmp, int x, int y) {
  for (; x <= cnt; x += (x & (-x))) c[tmp][x] += y;
}
int main() {
  scanf("%d %lld", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + n + 1);
  cnt = unique(b + 1, b + n + 1) - b - 1;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + cnt + 1, a[i]) - b;
  for (int i = 1; i <= n; i++) {
    sum += i - 1 - query(0, a[i]);
    add(0, a[i], 1);
  }
  int r = 1;
  for (int i = 1; i < n; i++) {
    sum += query(1, cnt) - query(1, a[i]) + query(0, a[i] - 1);
    add(1, a[i], 1);
    while (r <= i) {
      sum -= query(1, cnt) - query(1, a[r]) + query(0, a[r] - 1);
      add(0, a[r], -1);
      r++;
    }
    while (r <= n && sum > k) {
      sum -= query(1, cnt) - query(1, a[r]) + query(0, a[r] - 1);
      add(0, a[r], -1);
      r++;
    }
    ans += n + 1 - r;
  }
  printf("%lld", ans);
}
