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 int maxn = 1e5 + 21;
int st, n, m, a[maxn], suf[maxn], prf[maxn], Q[maxn];
long long k, cnt, ans;
void add(int x, int val, int* fen) {
  x++;
  while (x <= m) {
    fen[x] += val;
    x += (x & -x);
  }
}
int getf(int x, int* fen) {
  int ret = 0;
  while (x > 0) {
    ret += fen[x];
    x -= (x & -x);
  }
  return ret;
}
int get(int l, int r, int* fen) { return getf(r, fen) - getf(l, fen); }
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i], Q[m++] = a[i];
  sort(Q, Q + m);
  m = unique(Q, Q + m) - Q;
  for (int i = n - 1; i > 0; i--) {
    int ind = lower_bound(Q, Q + m, a[i]) - Q;
    cnt += get(0, ind, suf);
    add(ind, 1, suf);
  }
  st = 1;
  for (int i = 0; i < n; i++) {
    int ind = lower_bound(Q, Q + m, a[i]) - Q;
    cnt += get(ind + 1, m, prf);
    add(ind, 1, prf);
    cnt += get(0, ind, suf);
    while ((cnt > k || st <= i) && st < n) {
      ind = lower_bound(Q, Q + m, a[st]) - Q;
      add(ind, -1, suf);
      cnt -= get(0, ind, suf);
      cnt -= get(ind + 1, m, prf);
      st++;
    }
    ans += n - st;
  }
  cout << ans << "\n";
}
