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;
int chh;
const int N = 100005;
int n, e;
long long m;
int a[N << 2], b[N << 2], c[N], d[N];
int find(int a[], int k, int s, int t, int l, int r) {
  if (l > r) return 0;
  int mid = (s + t) >> 1;
  if (s >= l && t <= r) return a[k];
  if (s > r || t < l) return 0;
  return find(a, k << 1, s, mid, l, r) + find(a, k << 1 | 1, mid + 1, t, l, r);
}
void update(int a[], int k, int s, int t, int x, int val) {
  int mid = (s + t) >> 1;
  if (s == t)
    a[k] += val;
  else {
    if (x <= mid)
      update(a, k << 1, s, mid, x, val);
    else
      update(a, k << 1 | 1, mid + 1, t, x, val);
    a[k] = a[k << 1] + a[k << 1 | 1];
  }
}
int main() {
  int i, j, l, r;
  long long ans, tmp;
  while (cin >> n >> m) {
    e = 0;
    for (i = 1; i <= n; i++) {
      scanf("%d", &c[i]);
      d[e++] = c[i];
    }
    sort(d, d + e);
    e = unique(d, d + e) - d;
    for (i = 1; i <= n; i++) {
      c[i] = lower_bound(d, d + e, c[i]) - d + 1;
    }
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    ans = tmp = 0, l = 1, r = 2;
    update(a, 1, 1, e, c[1], 1);
    for (i = 2; i <= n; i++) {
      tmp += (long long)find(a, 1, 1, e, c[i] + 1, e);
      tmp += (long long)find(b, 1, 1, e, c[i] + 1, e);
      update(b, 1, 1, e, c[i], 1);
    }
    if (tmp <= m) {
      ans = (long long)n * (n - 1) / 2;
      cout << ans << "\n";
      continue;
    }
    while (1) {
      while (tmp > m && r < n) {
        tmp -= (long long)find(a, 1, 1, e, c[r] + 1, e);
        tmp -= (long long)find(b, 1, 1, e, 1, c[r] - 1);
        update(b, 1, 1, e, c[r], -1);
        r++;
      }
      if (tmp > m) break;
      ans += (long long)(n - r + 1);
      if (l == n - 1) break;
      l++;
      tmp += (long long)find(a, 1, 1, e, c[l] + 1, e);
      tmp += (long long)find(b, 1, 1, e, 1, c[l] - 1);
      update(a, 1, 1, e, c[l], 1);
    }
    cout << ans << '\n';
  }
  return 0;
}
