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>
const double eps = 1e-9;
const double pi = acos(-1);
const int oo = 1000000000;
const int mod = 1000000007;
const double E = 2.7182818284590452353602874713527;
using namespace std;
struct case1 {
  int v, p;
} p[100001];
int a[100001], tree[2][100001];
int o = 0;
bool cmp(case1 a, case1 b) { return a.v < b.v; }
void ins(int a, int i, int v) {
  for (; i <= o; i += ((i) & (-(i)))) tree[a][i] += v;
}
int sum(int a, int i) {
  int s = 0;
  for (; i; i -= ((i) & (-(i)))) s += tree[a][i];
  return s;
}
int main() {
  int n;
  long long k, tot = 0, ans = 0;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &p[i].v);
    p[i].p = i;
  }
  sort(p + 1, p + n + 1, cmp);
  for (int i = 1; i <= n; i++)
    a[p[i].p] = (o += (i == 1 || p[i].v > p[i - 1].v));
  int r = 2;
  for (int i = n; i >= 2; i--) {
    tot += sum(1, a[i] - 1);
    ins(1, a[i], 1);
  }
  tot += sum(1, a[1] - 1);
  ins(0, a[1], 1);
  for (int l = 1; l < n; l++) {
    while (r <= l || r <= n && tot > k) {
      tot -= sum(1, a[r] - 1) + l - sum(0, a[r]);
      ins(1, a[r], -1);
      r++;
    }
    ans += n - r + 1;
    tot += sum(1, a[l + 1] - 1) + l - sum(0, a[l + 1]);
    ins(0, a[l + 1], 1);
  }
  cout << ans << endl;
  return 0;
}
