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 lst[4 * 100005];
int rst[4 * 100005];
int arr[100005];
void update_tree(int ss, int se, int si, int *st, int ui, int val) {
  if (ss == se) {
    st[si] += val;
    return;
  }
  int mid = ss + (se - ss) / 2;
  if (ui <= mid)
    update_tree(ss, mid, 2 * si + 1, st, ui, val);
  else
    update_tree(mid + 1, se, 2 * si + 2, st, ui, val);
  st[si] = st[2 * si + 1] + st[2 * si + 2];
}
long long int query_tree(int ss, int se, int si, int *st, int qs, int qe) {
  if (qs > qe) return 0;
  if (ss > qe || se < qs) return 0;
  if (ss >= qs && se <= qe) return st[si];
  int mid = ss + (se - ss) / 2;
  long long int ans = query_tree(ss, mid, 2 * si + 1, st, qs, qe);
  ans += query_tree(mid + 1, se, 2 * si + 2, st, qs, qe);
  return ans;
}
int main() {
  pair<int, int> temp[100005];
  int n;
  long long int k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> temp[i].first;
    temp[i].second = i;
  }
  sort(temp, temp + n);
  int val = 0;
  arr[temp[0].second] = val;
  for (int i = 1; i < n; i++) {
    if (temp[i].first != temp[i - 1].first) val++;
    arr[temp[i].second] = val;
  }
  long long int l_inv = 0, r_inv = 0;
  for (int i = 0; i < n - 1; i++) {
    update_tree(0, n - 1, 0, lst, arr[i], 1);
    if (arr[i] < n - 1)
      l_inv += query_tree(0, n - 1, 0, lst, arr[i] + 1, n - 1);
  }
  long long int t_inv = l_inv;
  t_inv += query_tree(0, n - 1, 0, lst, arr[n - 1] + 1, n - 1);
  if (t_inv <= k) {
    long long int ans = n;
    ans *= (n - 1);
    ans /= 2;
    cout << ans;
    return 0;
  }
  int l_idx = n - 2;
  while (l_inv > k) {
    int to_rem = arr[l_idx];
    update_tree(0, n - 1, 0, lst, to_rem, -1);
    long long int rems = 0;
    if (to_rem < n - 1) {
      rems = query_tree(0, n - 1, 0, lst, to_rem + 1, n - 1);
    }
    l_inv -= rems;
    l_idx--;
  }
  long long int ans = 0;
  long long int c_inv = 0;
  int r_idx = n;
  while (r_idx > 1) {
    r_idx--;
    r_inv += query_tree(0, n - 1, 0, rst, 0, arr[r_idx] - 1);
    update_tree(0, n - 1, 0, rst, arr[r_idx], 1);
    c_inv += query_tree(0, n - 1, 0, lst, arr[r_idx] + 1, n - 1);
    while (l_inv + c_inv + r_inv > k) {
      if (l_idx < 0) break;
      int to_rem = arr[l_idx];
      l_idx--;
      update_tree(0, n - 1, 0, lst, to_rem, -1);
      long long int rems = 0;
      rems = query_tree(0, n - 1, 0, lst, to_rem + 1, n - 1);
      l_inv -= rems;
      c_inv -= query_tree(0, n - 1, 0, rst, 0, to_rem - 1);
    }
    if (l_idx >= 0) {
      ans += (l_idx + 1);
    }
  }
  cout << ans;
  return 0;
}
