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 s0[100010], s1[100010], sn;
inline int lowbit(int n) { return n & (-n); }
void add(int seg[], int x, int c) {
  for (int i = x; i <= sn; i += lowbit(i)) seg[i] += c;
}
int sum(int seg[], int x) {
  int i, ans = 0;
  for (i = x; i > 0; i &= i - 1) ans += seg[i];
  return ans;
}
int c[100010], a[100010];
map<int, int> mp;
int main() {
  int n, i, p;
  long long x, k;
  scanf("%d%I64d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &c[i]);
    a[i] = c[i];
  }
  sort(a, a + n);
  mp[a[0]] = 2;
  int m = 3;
  for (i = 1; i < n; i++) {
    if (a[i] != a[i - 1]) {
      mp[a[i]] = m++;
    }
  }
  for (i = 0; i < n; i++) {
    c[i] = mp[c[i]];
  }
  sn = m;
  p = 1;
  x = 0;
  add(s0, c[0], 1);
  for (i = 1; i < n; i++) {
    add(s1, c[i], 1);
    x += i - sum(s1, c[i]);
  }
  x += sum(s1, c[0] - 1);
  int n0 = 1, n1 = n - 1;
  while (p < n && x > k) {
    add(s1, c[p], -1);
    n1--;
    x -= n0 - sum(s0, c[p]);
    x -= sum(s1, c[p] - 1);
    p++;
  }
  long long ans = n - p;
  for (i = 1; i < n - 1; i++) {
    add(s0, c[i], 1);
    n0++;
    x += n0 - sum(s0, c[i]);
    x += sum(s1, c[i] - 1);
    while ((p < n && x > k) || (i == p)) {
      add(s1, c[p], -1);
      n1--;
      x -= n0 - sum(s0, c[p]);
      x -= sum(s1, c[p] - 1);
      p++;
    }
    ans += n - p;
  }
  printf("%I64d\n", ans);
  return 0;
}
