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 = 1e5 + 1e3;
int n, m, a[N], b[N], h[N], c2[N], s2, c[N], inv1[N], inv2[N], c1[N];
long long k;
long long ans, s;
inline int query(int* a, int x) {
  if (!x) return 0;
  int z(0);
  for (int i = x; i; i -= i & -i) z += a[i];
  return z;
}
inline void updata(int* a, int x, int v) {
  for (int i = x; i <= m; i += i & -i) a[i] += v;
}
int main() {
  scanf("%d", &n);
  cin >> k;
  for (int i = 0; i < n; i++) scanf("%d", &a[i]), b[i] = a[i];
  m = n;
  sort(b, b + m);
  m = unique(b, b + m) - b;
  for (int i = 0; i < n; i++) h[i] = lower_bound(b, b + m, a[i]) - b + 1;
  for (int i = 0; i < n; i++)
    inv1[i] = query(c, m) - query(c, h[i]), updata(c, h[i], 1);
  memset(c, 0, sizeof c);
  for (int i = n - 1; i >= 0; i--)
    inv2[i] = query(c, h[i] - 1), updata(c, h[i], 1);
  for (int i = 1; i < n; i++) s += inv2[i], updata(c2, h[i], 1);
  s += inv2[0];
  updata(c1, h[0], 1);
  int l(0), r(1);
  for (; r < n; r++) {
    l++;
    while (l < r) {
      int t(0);
      t += inv1[l];
      t += query(c2, h[l] - 1);
      if (s + t <= k)
        updata(c1, h[l], 1), l++, s += t;
      else
        break;
    }
    if (s <= k) ans += l;
    l--;
    s -= inv2[r];
    s -= query(c1, m) - query(c1, h[r]);
    updata(c2, h[r], -1);
  }
  cout << ans;
  return 0;
}
