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 = 2e5 + 1000;
const int maxlog = 18;
long long fen[maxn][5], a[maxn], b[maxn];
void add(int x, long long val, int id) {
  for (x++; x < maxn; x += x & -x) {
    fen[x][id] += val;
  }
}
long long sum(int x, int id) {
  long long res = 0;
  for (x++; x > 0; x -= x & -x) {
    res += fen[x][id];
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
  }
  copy(a, a + n, b);
  sort(b, b + n);
  for (int i = 0; i < n; ++i) {
    a[i] = lower_bound(b, b + n, a[i]) - b;
  }
  long long inver = 0;
  for (int i = n - 1; i >= 0; --i) {
    inver += sum(a[i] - 1, 0);
    add(a[i], 1, 0);
  }
  long long res = 0, left = 0, right = 0;
  while (left < n) {
    inver = inver + sum(a[left] - 1, 0) + sum(n + 1, 2) - sum(a[left], 2);
    add(a[left], 1, 2);
    while ((right == left) || (inver > k && right != n)) {
      inver = inver - sum(a[right] - 1, 0) - sum(n + 1, 2) + sum(a[right], 2);
      add(a[right], -1, 0);
      ++right;
    }
    res += n - right;
    ++left;
  }
  cout << res << endl;
  return 0;
}
