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;
long long k, n;
void red(vector<int> &in) {
  vector<int> tmp = in;
  map<int, int> mm;
  sort(tmp.begin(), tmp.end());
  tmp.resize((tmp.begin(), tmp.end()) - tmp.begin());
  for (int i = (0); i < (tmp.size()); i += (1)) mm[tmp[i]] = i + 1;
  for (int i = (0); i < (in.size()); i += (1)) in[i] = mm[in[i]];
}
long long gg(long long sg[], int node, int s, int e, int i, int j) {
  if (s > e || s > j || e < i) return 0;
  if (s >= i && e <= j) return sg[node];
  return gg(sg, (node << 1), s, ((s + e) >> 1), i, j) +
         gg(sg, ((node << 1) | 1), ((s + e) >> 1) + 1, e, i, j);
}
void upd(long long sg[], int node, int s, int e, int i, int val) {
  if (s > e || s > i || e < i) return;
  if (s >= i && e <= i) {
    sg[node] += val;
    return;
  }
  upd(sg, (node << 1), s, ((s + e) >> 1), i, val);
  upd(sg, ((node << 1) | 1), ((s + e) >> 1) + 1, e, i, val);
  sg[node] = sg[(node << 1)] + sg[((node << 1) | 1)];
}
long long sg1[100010 * 4] = {0}, sg2[100010 * 4] = {0};
long long solve(vector<int> &in) {
  long long res = 0;
  long long curinv = 0;
  upd(sg1, 1, 1, n, in[0], 1);
  for (int i = (1); i < (n); i += (1)) {
    upd(sg2, 1, 1, n, in[i], 1);
    curinv += gg(sg2, 1, 1, n, in[i] + 1, n);
    curinv += gg(sg1, 1, 1, n, in[i] + 1, n);
  }
  res += (curinv <= k);
  int l = 0;
  for (int r = (2); r < (n); r += (1)) {
    upd(sg2, 1, 1, n, in[r - 1], -1);
    curinv -= gg(sg1, 1, 1, n, in[r - 1] + 1, n);
    curinv -= gg(sg2, 1, 1, n, 1, in[r - 1] - 1);
    l++;
    while (l < r) {
      long long tmp = 0;
      tmp += gg(sg1, 1, 1, n, in[l] + 1, n);
      tmp += gg(sg2, 1, 1, n, 1, in[l] - 1);
      if (tmp + curinv <= k)
        curinv += tmp;
      else
        break;
      upd(sg1, 1, 1, n, in[l], 1);
      l++;
    }
    l--;
    if (curinv <= k) res += l + 1;
  }
  return res;
}
int main() {
  cin >> n >> k;
  vector<int> in(n);
  for (int i = (0); i < (n); i += (1)) cin >> in[i];
  if (n == 1) {
    cout << 1;
    return 0;
  }
  red(in);
  cout << solve(in) << endl;
  return 0;
}
