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 INF = 1e9 + 10, MAX = 2e5 + 1e4, MOD = 1e9 + 7, MAXL = 25;
long long a[MAX], dpr[MAX];
int seg[4][4 * MAX];
pair<long long, long long> b[MAX];
void upd(int s, int id, int b, int e, int x, int y) {
  if (e - b == 1) {
    seg[s][id] += y;
    return;
  }
  int m = (b + e) / 2;
  if (x < m)
    upd(s, id * 2, b, m, x, y);
  else
    upd(s, id * 2 + 1, m, e, x, y);
  seg[s][id] = seg[s][id * 2] + seg[s][id * 2 + 1];
  return;
}
int g(int s, int id, int b, int e, int l, int r) {
  if (e <= l || r <= b) return 0;
  if (l <= b && e <= r) return seg[s][id];
  int m = (b + e) / 2;
  return g(s, id * 2, b, m, l, r) + g(s, id * 2 + 1, m, e, l, r);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k, ans = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = {a[i], i};
  }
  sort(b + 1, b + n + 1);
  long long c = 0;
  for (int i = 1; i <= n; i++) {
    if (b[i].first != b[i - 1].first) c++;
    a[b[i].second] = c;
  }
  for (int i = n; i > 1; i--) {
    dpr[i] = dpr[i + 1] + g(1, 1, 1, 1e5 + 10, 1, a[i]);
    upd(1, 1, 1, 1e5 + 10, a[i], 1);
  }
  long long r = 2, cl = 0;
  for (int i = 1; i < n; i++) {
    cl += g(0, 1, 1, 1e5 + 10, a[i] + 1, 1e5 + 10);
    upd(0, 1, 1, 1e5 + 10, a[i], 1);
    cl += g(1, 1, 1, 1e5 + 10, 1, a[i]);
    while ((cl + dpr[r] > k || r <= i) && r <= n) {
      cl -= g(0, 1, 1, 1e5 + 10, a[r] + 1, 1e5 + 10);
      upd(1, 1, 1, 1e5 + 10, a[r], -1);
      r++;
    }
    ans += n + 1 - r;
  }
  cout << ans;
  return 0;
}
