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 = 100010;
int a[N], n;
long long m;
class BIT {
 public:
  int a[N];
  void add(int x, int y) {
    for (int i = x; i < N; a[i] += y, i += i & -i)
      ;
  }
  int getS(int x) {
    int ans = 0;
    for (int i = x; i; ans += a[i], i -= i & -i)
      ;
    return ans;
  }
  void clear() { memset(a, 0, N * sizeof(int)); }
};
BIT A, B;
long long dl[N], dr[N];
int h[N], q;
int main() {
  int i, j, k;
  cin >> n >> m;
  for (i = 1; i <= n; ++i) {
    cin >> a[i];
    h[i] = a[i];
  }
  sort(h + 1, h + n + 1);
  q = unique(h + 1, h + n + 1) - h;
  for (i = 1; i <= n; ++i) a[i] = lower_bound(h + 1, h + q, a[i]) - h;
  A.clear();
  B.clear();
  long long cnt = 0, ans = 0;
  for (i = 1; i <= n; ++i) {
    cnt += (i - 1) - A.getS(a[i]);
    A.add(a[i], 1);
  }
  A.add(a[n], -1);
  B.add(a[n], 1);
  if (cnt <= m) {
    cout << ((long long)n) * (n - 1) / 2 << endl;
    return 0;
  }
  i = n - 1, j = n;
  while (j > i && i >= 1) {
    while (cnt > m && i > 1) {
      A.add(a[i], -1);
      cnt -= (i - 1) - A.getS(a[i]) + B.getS(a[i] - 1);
      --i;
    }
    if (cnt > m) break;
    ans += i;
    while (cnt <= m && j > i + 1) {
      j--;
      cnt += B.getS(a[j] - 1) + i - A.getS(a[j]);
      B.add(a[j], 1);
      if (cnt <= m) ans += i;
    }
  }
  cout << ans << endl;
  return 0;
}
