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 MAX = 2e5 + 3, inf = 1e18 + 9, mod = 1e9 + 7;
int n, arr[MAX], b[MAX];
long long k, fen[MAX][2], inv;
void add(int ind, int val, bool t) {
  for (int x = ind + 1; x < MAX; x += x & -x) fen[x][t] += val;
}
long long get(int ind, bool t) {
  long long res = 0;
  for (int x = ind + 1; x > 0; x -= x & -x) res += fen[x][t];
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> arr[i], b[i] = arr[i];
  sort(b, b + n);
  for (int i = 0; i < n; i++) arr[i] = lower_bound(b, b + n, arr[i]) - b;
  for (int i = n - 1; i >= 1; i--) {
    inv += get(arr[i] - 1, 1);
    add(arr[i], 1, 1);
  }
  int l = 0, r = 1;
  long long ans = 0;
  while (l < n) {
    inv += get(arr[l] - 1, 1);
    inv += get(MAX - 2, 0) - get(arr[l], 0);
    add(arr[l], 1, 0);
    while (r <= l || (inv > k && r < n)) {
      inv -= get(MAX - 2, 0) - get(arr[r], 0);
      inv -= get(arr[r] - 1, 1);
      add(arr[r], -1, 1);
      r++;
    }
    ans += n - r;
    l++;
  }
  cout << ans << endl;
  return 0;
}
