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 + 100, inF = 2e9;
long long n, k, now, ans, a[N];
vector<pair<int, int> > seg[N << 2];
map<int, int> cnt;
void build(int v = 1, int s = 0, int e = n) {
  cnt.clear();
  for (int i = s; i < e; i++) cnt[a[i]]++;
  seg[v].push_back(pair<int, int>(0, 0));
  for (auto p : cnt)
    seg[v].push_back(pair<int, int>(p.first, p.second + seg[v].back().second));
  if (e - s == 1) return;
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  build(lc, s, mid);
  build(rc, mid, e);
}
int get(int l, int r, int x, int v = 1, int s = 0, int e = n) {
  if (r <= s || e <= l) return 0;
  if (l <= s && e <= r)
    return (*(--upper_bound(seg[v].begin(), seg[v].end(),
                            pair<int, int>(x, inF))))
        .second;
  int mid = (s + e) >> 1, lc = v << 1, rc = lc | 1;
  return get(l, r, x, lc, s, mid) + get(l, r, x, rc, mid, e);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  build();
  for (int i = 0; i < n - 1; i++) now += get(i + 1, n, a[i] - 1);
  if (now <= k) ans += n - 1;
  int r = 0;
  for (int i = 0; i < n - 1; i++) {
    if (i) {
      now += i - get(0, i, a[i - 1]);
      now += (r == n ? 0 : get(r, n, a[i - 1] - 1));
    }
    while ((now > k && r < n - 1) || r < i + 1) {
      now -= (i == 0 ? 0 : i - (get(0, i, a[r])));
      now -= get(r, n, a[r] - 1);
      r++;
    }
    if (now <= k && i) ans += n - r;
  }
  cout << ans;
  return 0;
}
