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 MAX = 1e5 + 1e2;
int A[MAX], B[MAX], n, m;
int BIT[2][MAX];
int pref[MAX], suff[MAX];
void update(int p, int v, int *T) {
  for (; p <= m + 1; p += ((p) & (-(p)))) T[p] += v;
}
int get(int p, int *T) {
  int ret = 0;
  for (; p; p ^= ((p) & (-(p)))) ret += T[p];
  return ret;
}
int main() {
  long long k;
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < (n); i++) scanf("%d", A + i), B[i] = A[i];
  sort(B, B + n);
  m = unique(B, B + n) - B;
  for (int i = 0; i < (n); i++) A[i] = upper_bound(B, B + m, A[i]) - B;
  long long inv = 0, tot_inv = 0, ans = 0, cnt = 0, inv_out;
  for (int i = 0; i < (n); i++) {
    tot_inv += get(m + 1, BIT[0]) - get(A[i], BIT[0]);
    update(A[i], 1, BIT[0]);
  }
  inv_out = tot_inv;
  memset(BIT, 0, sizeof BIT);
  for (int i = n - 1; i >= 0; i--) {
    suff[i] = get(A[i] - 1, BIT[0]);
    update(A[i], 1, BIT[0]);
  }
  memset(BIT, 0, sizeof BIT);
  int l = 0;
  update(A[l], 1, BIT[1]);
  for (int r = 1; r < n; r++) {
    if (r != l + 1) {
      inv += get(m + 1, BIT[0]) - get(A[r - 1], BIT[0]);
      inv_out -= suff[r - 1];
      inv_out -= get(m + 1, BIT[1]) - get(A[r - 1], BIT[1]);
      update(A[r - 1], 1, BIT[0]);
      assert(inv >= 0);
      assert(inv_out >= 0);
    }
    while (l + 1 < r && inv_out <= k) {
      ++l;
      inv -= get(A[l] - 1, BIT[0]);
      update(A[l], -1, BIT[0]);
      inv_out += suff[l] - get(A[l] - 1, BIT[0]);
      inv_out += get(m + 1, BIT[1]) - get(A[l], BIT[1]);
      update(A[l], 1, BIT[1]);
    }
    assert(inv >= 0);
    assert(inv_out >= 0);
    ans += l + (inv_out <= k);
  }
  cerr << ans << endl;
  cout << ans << endl;
  return 0;
}
