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 tree[3][100009], a[100009], n = 100004;
void update(int idx, int val, int code) {
  if (idx == 0) return;
  while (idx <= n) {
    tree[code][idx] += val;
    idx += (idx & -idx);
  }
}
int sum(int idx, int code) {
  int val = 0;
  while (idx > 0) {
    val += tree[code][idx];
    idx -= (idx & -idx);
  }
  return val;
}
int main() {
  vector<int> tmp;
  int n;
  long long int k;
  scanf("%d", &n);
  scanf("%lld", &k);
  map<int, int> idx;
  for (int i = int(0); i <= int(n - 1); i++) {
    scanf("%d", &a[i]);
    tmp.push_back(a[i]);
  }
  sort(tmp.begin(), tmp.end());
  for (int i = int(0); i <= int(n - 1); i++) idx[tmp[i]] = i + 1;
  long long int inv = 0, ans = 0;
  for (int i = int(1); i <= int(n - 1); i++) {
    update(idx[a[i]], 1, 1);
    inv += sum(100002, 1) - sum(idx[a[i]], 1);
  }
  int left = -1, right = 1;
  while (right < n) {
    left++;
    if (left == right) {
      inv -= (sum(idx[a[right]] - 1, 1));
      inv -= (sum(100002, 0) - sum(idx[a[right]], 0));
      update(idx[a[right]], -1, 1);
      right++;
    }
    inv += (sum(idx[a[left]] - 1, 1));
    inv += (sum(100002, 0) - sum(idx[a[left]], 0));
    update(idx[a[left]], 1, 0);
    while (inv > k && right < n) {
      inv -= (sum(idx[a[right]] - 1, 1));
      inv -= (sum(100002, 0) - sum(idx[a[right]], 0));
      update(idx[a[right]], -1, 1);
      right++;
    }
    ans += n - right;
  }
  printf("%lld\n", ans);
  return 0;
}
