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 double eps = 1E-8;
const int dx4[4] = {1, 0, 0, -1};
const int dy4[4] = {0, -1, 1, 0};
const int inf = 0x3f3f3f3f;
const int N = 2E5 + 7;
int tree[2][N];
int a[N], p[N];
int n;
long long k;
int lowbit(int x) { return x & (-x); }
void update(int o, int x, int delta) {
  if (!o) x = n - x + 1;
  for (int i = x; i <= n; i += lowbit(i)) tree[o][i] += delta;
}
int Hash(int x) { return lower_bound(p + 1, p + n + 1, x) - p; }
int Sum(int o, int x) {
  if (!o) x = n - x + 1;
  int res = 0;
  for (int i = x; i >= 1; i -= lowbit(i)) res += tree[o][i];
  return res;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i], p[i] = a[i];
  sort(p + 1, p + n + 1);
  long long total = 0;
  for (int i = n; i >= 1; i--) {
    int x = Hash(a[i]);
    a[i] = x;
    total += Sum(1, a[i] - 1);
    update(1, a[i], 1);
  }
  int r = 1;
  long long ans = 0;
  for (int l = 1; l <= n; l++) {
    total += Sum(0, a[l] + 1) + Sum(1, a[l] - 1);
    update(0, a[l], 1);
    while ((r <= l || total > k) && r <= n) {
      total -= Sum(0, a[r] + 1) + Sum(1, a[r] - 1);
      update(1, a[r], -1);
      r++;
    }
    ans += n - r + 1;
  }
  cout << ans << endl;
  return 0;
}
