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 = 2e5 + 3;
int n, m, i, j, l, r, d, u, ii, jj, h[N];
long long first[N][3], k, t, hh[N], x, y, z, s;
map<int, int> MP;
long long f(int x, int d) {
  long long t = 0;
  for (; x; x -= x & -x) t += first[x][d];
  return t;
}
void up(int x, int d, int k) {
  for (; x and x <= s; x += x & -x) first[x][d] += k;
}
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) {
    scanf("%d", h + i);
    hh[i] = h[i];
  }
  sort(hh + 1, hh + 1 + n);
  for (i = 1; i <= n; i++)
    if (hh[i] != hh[i - 1]) MP[hh[i]] = ++s;
  for (i = n; i; i--) {
    hh[i] = hh[i + 1] + f(MP[h[i]] - 1, 0);
    up(MP[h[i]], 0, 1);
  }
  r = 1;
  for (i = 1; i < n; i++) {
    up(MP[h[i]], 1, 1);
    while (1) {
      x = hh[1] - y - z;
      if (i == r)
        r++;
      else if (x <= k) {
        if (r <= n) t += n - r + 1;
        if (r - i > 1) {
          y -= hh[i + 1] - hh[i + 2] - f(MP[h[i + 1]] - 1, 2);
          z -= f(s, 1) - f(MP[h[i + 1]], 1);
          up(MP[h[i + 1]], 2, -1);
        }
        x = hh[1] - y - z;
        break;
      } else {
        y += hh[r] - hh[r + 1];
        z += f(s, 1) - f(MP[h[r]], 1);
        up(MP[h[r]], 2, 1);
        r++;
        x = hh[1] - y - z;
      }
    }
  }
  cout << t << '\n';
}
