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 n;
int a[100005];
vector<int> fenwick[100005];
void build() {
  for (int i = 1; i <= n; ++i) {
    for (int j = i; j > i - (i & (-i)); j--) fenwick[i].push_back(a[j]);
    sort(fenwick[i].begin(), fenwick[i].end());
  }
}
long long int solve_g(int idx, int x) {
  long long int ans = 0;
  for (int i = idx; i; i -= i & (-i)) {
    ans += fenwick[i].size() -
           (upper_bound(fenwick[i].begin(), fenwick[i].end(), x) -
            fenwick[i].begin());
  }
  return ans;
}
long long int solve_l(int idx, int x) {
  long long int ans = 0;
  for (int i = idx; i; i -= i & (-i)) {
    ans += (upper_bound(fenwick[i].begin(), fenwick[i].end(), x - 1) -
            fenwick[i].begin());
    ;
  }
  return ans;
}
int k;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  build();
  int l = 1, r = 2;
  long long int curr = 0;
  for (int i = n; i > 1; i--) {
    curr += solve_g(i - 1, a[i]);
  }
  long long int ans = 0;
  if (k >= curr) {
    ans = n;
    ans *= (n - 1);
    ans /= 2;
    cout << ans << endl;
    return 0;
  }
  while (r <= n) {
    if (curr > k) {
      r++;
      curr -= solve_g(l, a[r - 1]) + solve_l(n, a[r - 1]) -
              solve_l(r - 1, a[r - 1]);
    } else {
      l++;
      curr += solve_g(l - 1, a[l]) + solve_l(n, a[l]) - solve_l(r - 1, a[l]);
    }
    if (curr <= k) ans += n + 1 - r;
  }
  cout << ans << endl;
  return 0;
}
