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 = 100 * 1000 + 5;
int a[MAXN], st[2][4 * MAXN];
map<int, int> mp;
void add(int t, int v, int tl, int tr, int pos, int val) {
  if (tl == tr - 1) {
    st[t][v] += val;
    return;
  }
  int tm = (tl + tr) / 2;
  if (pos < tm)
    add(t, 2 * v, tl, tm, pos, val);
  else
    add(t, 2 * v + 1, tm, tr, pos, val);
  st[t][v] = st[t][2 * v] + st[t][2 * v + 1];
}
int get(int t, int v, int tl, int tr, int l, int r) {
  if (l == tl && r == tr) return st[t][v];
  int tm = (tl + tr) / 2, res = 0;
  if (l < tm) res += get(t, 2 * v, tl, tm, l, min(r, tm));
  if (r > tm) res += get(t, 2 * v + 1, tm, tr, max(l, tm), r);
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]];
  }
  int m = 0;
  for (map<int, int>::iterator it = mp.begin(); it != mp.end(); it++)
    it->second = m++;
  for (int i = 0; i < n; i++) a[i] = mp[a[i]];
  long long cnt = 0;
  for (int i = n - 1; i > 0; i--) {
    if (a[i] > 0) cnt += get(1, 1, 0, m, 0, a[i]);
    add(1, 1, 0, m, a[i], 1);
  }
  long long ans = 0;
  for (int i = 0, j = 1; i < n - 1; i++) {
    if (j == i) {
      if (a[j] < m - 1) cnt -= get(0, 1, 0, m, a[j] + 1, m);
      if (a[j] > 0) cnt -= get(1, 1, 0, m, 0, a[j]);
      add(1, 1, 0, m, a[j], -1);
      j++;
    }
    if (a[i] < m - 1) cnt += get(0, 1, 0, m, a[i] + 1, m);
    if (a[i] > 0) cnt += get(1, 1, 0, m, 0, a[i]);
    add(0, 1, 0, m, a[i], 1);
    while (j < n && cnt > k) {
      if (a[j] < m - 1) cnt -= get(0, 1, 0, m, a[j] + 1, m);
      if (a[j] > 0) cnt -= get(1, 1, 0, m, 0, a[j]);
      add(1, 1, 0, m, a[j], -1);
      j++;
    }
    ans += n - j;
  }
  cout << ans << '\n';
  return 0;
}
