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 maxn = 1e5 + 100;
long long int n, inv, ans;
int a[maxn], fen[maxn][2];
void add(int idx, int val, int dr) {
  for (idx++; idx < maxn; idx += idx & -idx) fen[idx][dr] += val;
}
int get(int idx, int dr) {
  int sum = 0;
  for (; idx > 0; idx -= idx & -idx) sum += fen[idx][dr];
  return sum;
}
int get(int l, int r, int dr) { return get(r, dr) - get(l, dr); }
void _add(int idx) {
  add(a[idx], 1, 0);
  inv += get(0, a[idx], 1);
  inv += get(a[idx] + 1, n, 0);
}
void _rem(int idx) {
  add(a[idx], -1, 1);
  inv -= get(a[idx] + 1, n, 0);
  inv -= get(0, a[idx], 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<int> v;
  for (int i = 0; i < n; i++) v.push_back(a[i]);
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin();
  for (int i = 0; i < n; i++) {
    add(a[i], 1, 1);
    inv += get(a[i] + 1, n, 1);
  }
  int r = 0;
  for (int l = 0; l < n; l++) {
    _add(l);
    if (l == r) _rem(r++);
    while (r < n && inv > k) _rem(r++);
    ans += n - r;
  }
  cout << ans << endl;
  return 0;
}
