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 = 1e5 + 2;
int n, k, a[N];
int fen[3][N];
long long ans, kol;
vector<int> vec;
void update(int pos, int val, int type) {
  for (; pos <= n; pos |= pos + 1) fen[type][pos] += val;
}
int get(int r, int type) {
  int res = 0;
  for (; r > 0; r = (r & (r + 1)) - 1) res += fen[type][r];
  return res;
}
int get(int l, int r, int type) { return get(r, type) - get(l - 1, type); }
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    vec.push_back(a[i]);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin() + 1;
  for (int i = n; i >= 2; i--) {
    kol += get(1, a[i] - 1, 2);
    update(a[i], 1, 2);
  }
  int r = 2;
  for (int i = 1; i < n; i++) {
    kol += get(a[i] + 1, n, 1) + get(1, a[i] - 1, 2);
    update(a[i], 1, 1);
    while (r <= n && (kol > k || r <= i)) {
      kol -= get(a[r] + 1, n, 1) + get(1, a[r] - 1, 2);
      update(a[r], -1, 2);
      r++;
    }
    ans += n - r + 1;
  }
  cout << ans << endl;
}
