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 Mn = 100010;
int n, m;
long long k;
int a[Mn], b[Mn], p[Mn];
long long sl[Mn], sr[Mn];
long long tl[Mn], tr[Mn];
inline void updatal(int x, int y) {
  for (; x <= m; x += x & -x) sl[x] += y;
}
inline long long getl(int x) {
  long long res(0);
  for (; x > 0; x -= x & -x) res += sl[x];
  return res;
}
inline void updatar(int x, int y) {
  for (; x <= m; x += x & -x) sr[x] += y;
}
inline long long getr(int x) {
  long long res(0);
  for (; x > 0; x -= x & -x) res += sr[x];
  return res;
}
int main() {
  scanf("%d%I64d", &m, &k);
  n = m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
  sort(b + 1, b + n + 1);
  m = unique(b + 1, b + m + 1) - b - 1;
  for (int i = 1; i <= n; i++) p[i] = lower_bound(b + 1, b + m + 1, a[i]) - b;
  int l, r;
  l = 1;
  r = 2;
  for (int i = n; i >= 2; i--) {
    tr[i] = getr(p[i] - 1) + tr[i + 1];
    updatar(p[i], 1);
  }
  memset(sl, 0, sizeof sl);
  long long mid(0);
  long long ans(0);
  for (int l = 1; l < n; l++) {
    tl[l] = l - getl(p[l]) - 1 + tl[l - 1];
    mid += getr(p[l] - 1);
    updatal(p[l], 1);
    while ((tl[l] + tr[r] + mid > k || l >= r) && r <= n) {
      updatar(p[r], -1);
      mid -= (l - getl(p[r]));
      r++;
    }
    if (r > n) break;
    if (tl[l] + tr[r] + mid <= k && l < r) ans += (n - r + 1);
  }
  cout << ans << endl;
  return 0;
}
