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>
const int sz = 100200, tn = 1073741824;
int l[sz * 66], r[sz * 66], s[sz * 66], N;
int ins(int x, int a, int S) {
  int m = ++N;
  s[m] = s[x] + 1;
  if (S == 1)
    return m;
  else if (a < (S >> 1)) {
    l[m] = ins(l[x], a, S >> 1);
    r[m] = r[x];
  } else {
    l[m] = l[x];
    r[m] = ins(r[x], a - (S >> 1), S >> 1);
  }
  return m;
}
int sum(int x, int a, int S) {
  if (S == 1)
    return s[x];
  else if (a < (S >> 1))
    return sum(l[x], a, S >> 1);
  else
    return sum(r[x], a - (S >> 1), S >> 1) + s[l[x]];
}
int n, a[sz], rtp[sz], rts[sz];
long long int k, inv = 0, ans = 0;
int L, R;
void delr() {
  inv -= (L - sum(rtp[L], a[R], tn));
  inv -= sum(rts[R + 1], a[R] - 1, tn);
  R++;
}
void inslplus1() {
  inv += (L - sum(rtp[L], a[L + 1], tn));
  inv += sum(rts[R], a[L + 1] - 1, tn);
  L++;
}
int main() {
  int i;
  scanf("%d %I64d", &n, &k);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (i = 1; i <= n; i++) rtp[i] = ins(rtp[i - 1], a[i], tn);
  for (i = n; i; i--) rts[i] = ins(rts[i + 1], a[i], tn);
  inv = 0;
  for (i = 1; i <= n; i++) inv += sum(rts[i + 1], a[i] - 1, tn);
  for (L = 1, R = 2; L <= n;) {
    while (R <= n && inv > k) delr();
    ans += n - R + 1;
    if (R == L + 1) delr();
    inslplus1();
  }
  printf("%I64d", ans);
  return 0;
}
