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 = (int)1e5 + 7;
const int inf = (int)1e9 + 7;
struct fenwick {
  int tree[N];
  void upd(int pos, int val) {
    for (; pos < N; pos += pos & -pos) {
      tree[pos] += val;
    }
  }
  int get(int r) {
    int res = 0;
    for (; r > 0; r -= r & -r) {
      res += tree[r];
    }
    return res;
  }
  int less(int r) { return get(r - 1); }
  int greater(int r) { return get(N - 1) - get(r); }
} l, r;
pair<int, int> ar[N];
int a[N];
int main() {
  int n;
  long long k;
  scanf("%d %I64d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ar[i] = make_pair(a[i], i);
  }
  sort(ar + 1, ar + n + 1);
  int j = 0;
  for (int i = 1; i <= n; i++) {
    if (ar[i].first != ar[i - 1].first) j++;
    a[ar[i].second] = j;
  }
  long long res = 0;
  for (int i = n; i >= 2; i--) {
    res += r.less(a[i]);
    r.upd(a[i], 1);
  }
  int right = 2;
  long long ans = 0;
  for (int i = 1; i < n; i++) {
    res += l.greater(a[i]) + r.less(a[i]);
    l.upd(a[i], 1);
    while (right <= n && (res > k || right <= i)) {
      res -= (l.greater(a[right]) + r.less(a[right]));
      r.upd(a[right], -1);
      right++;
    }
    ans += (n - right + 1);
  }
  printf("%I64d", ans);
}
