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 maxN = 100000 + 10, maxT = maxN * 5;
int n, a[maxN];
int b[maxN], nb, d[maxN];
long long k;
void enter() {
  scanf("%d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i - 1] = a[i];
  sort(b, b + n);
  nb = unique(b, b + n) - b;
  for (int i = 1; i <= n; i++) a[i] = lower_bound(b, b + nb, a[i]) - b + 1;
}
void update_d(int i, int val) {
  for (; i > 0; i -= i & -i) d[i] += val;
}
int get_d(int i) {
  int res = 0;
  for (; i < maxN; i += i & -i) res += d[i];
  return res;
}
void remove(long long &cc, int t) {
  cc -= get_d(a[t] + 1);
  for (int j = a[t] - 1; j > 0; j -= j & -j) cc -= b[j];
  update_d(a[t], -1);
}
void solve() {
  memset(b, 0, sizeof b);
  memset(d, 0, sizeof d);
  long long res = 0, cc = 0;
  for (int i = 1; i < n; i++) {
    update_d(a[i], 1);
    cc += get_d(a[i] + 1);
  }
  int t = n - 1;
  for (int i = n; i >= 1; i--) {
    for (int j = a[i] - 1; j > 0; j -= j & -j) cc += b[j];
    for (int j = a[i]; j < maxN; j += j & -j) b[j]++;
    cc += get_d(a[i] + 1);
    while (cc > k && t > 0) {
      remove(cc, t), t--;
    }
    res += t;
    if (t > 0 && t == i - 1) remove(cc, t), t--;
  }
  printf("%I64d\n", res);
}
int main() {
  enter();
  solve();
  return 0;
}
