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 mx = 112345;
const int BIT_U = 110000;
int v[mx], u[mx];
int A[mx], B[mx];
void upd(int *bit, int idx, int val) {
  idx++;
  while (idx < mx) {
    bit[idx] += val;
    idx += idx & -idx;
  }
}
int query(int *bit, int idx) {
  idx++;
  int sum = 0;
  while (idx) {
    sum += bit[idx];
    idx -= idx & -idx;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    u[i] = v[i];
  }
  sort(u, u + n);
  for (int i = 0; i < n; i++) {
    v[i] = lower_bound(u, u + n, v[i]) - u;
  }
  long long inv = 0;
  for (int i = 0; i < n - 1; i++) {
    upd(A, v[i], +1);
    inv += query(A, BIT_U) - query(A, v[i]);
  }
  inv += query(A, BIT_U) - query(A, v[n - 1]);
  upd(B, v[n - 1], +1);
  long long ans = 0;
  int r, l;
  r = n - 1;
  l = r - 1;
  while (l >= 0) {
    while (inv > k && l >= 0) {
      inv -= query(B, v[l] - 1);
      inv -= query(A, BIT_U) - query(A, v[l]);
      upd(A, v[l], -1);
      l--;
    }
    if (l < 0) continue;
    ans += l + 1;
    if (l + 1 == r) {
      inv -= query(B, v[l] - 1);
      inv -= query(A, BIT_U) - query(A, v[l]);
      upd(A, v[l], -1);
      l--;
    }
    r--;
    upd(B, v[r], +1);
    inv += query(B, v[r] - 1);
    inv += query(A, BIT_U) - query(A, v[r]);
  }
  cout << ans << endl;
}
