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 N = 1000 * 100 + 10;
int n, a[N];
long long k;
struct FenwickTree {
  vector<int> fen;
  FenwickTree(int n = 0) { fen.resize(n + 1, 0); }
  void upd(int x, int w = +1) {
    for (; x <= n; x += x & (-x)) fen[x] += w;
  }
  int get(int x) {
    int res = 0;
    for (; x; x -= x & (-x)) res += fen[x];
    return res;
  }
  int get(int l, int r) { return get(r) - get(l - 1); }
} prv, nxt;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> k;
  vector<int> all;
  for (int i = 1; i <= n; i++) cin >> a[i], all.push_back(a[i]);
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(all.begin(), all.end(), a[i]) - all.begin() + 1;
  long long inv = 0, ans = 0;
  nxt = prv = FenwickTree(n);
  for (int i = n; i >= 2; i--) {
    inv += nxt.get(a[i] - 1);
    nxt.upd(a[i]);
  }
  int r = 2;
  for (int l = 1; l < n; l++) {
    inv += nxt.get(a[l] - 1);
    inv += prv.get(a[l] + 1, n);
    prv.upd(a[l]);
    while (inv > k && r <= n) {
      nxt.upd(a[r], -1);
      inv -= nxt.get(a[r] - 1);
      inv -= prv.get(a[r] + 1, n);
      r++;
    }
    if (l == r) return cout << 1LL * n * (n - 1) / 2 << '\n', 0;
    ans += n - r + 1;
  }
  cout << ans << '\n';
  return 0;
}
