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 nMax = 100010;
int a[nMax], b[nMax], s[nMax * 3], s2[nMax * 3];
void Modify(int* c, int x, int y) {
  while (x < nMax * 3 - 1) {
    c[x] += y;
    x += x & (-x);
  }
}
int getsum(int* c, int x) {
  int res = 0;
  while (x) {
    res += c[x];
    x -= x & (-x);
  }
  return res;
}
int main() {
  long long sum, k, ans;
  int n, m, ta, he, i;
  while (scanf("%d%I64d", &n, &k) != EOF) {
    m = 0;
    sum = 0;
    ans = 0;
    memset(s, 0, sizeof(s));
    memset(s2, 0, sizeof(s2));
    for (i = 1; i <= n; i++) scanf("%d", a + i), b[m++] = a[i];
    sort(b, b + m);
    m = unique(b, b + m) - b;
    for (i = 1; i <= n; i++) a[i] = lower_bound(b, b + m, a[i]) - b + 1;
    for (i = 1; i <= n; i++) {
      Modify(s, a[i], 1);
      sum += i - getsum(s, a[i]);
    }
    if (sum <= k) {
      printf("%I64d\n", (long long)n * (n - 1) / 2);
      continue;
    }
    he = n - 1, ta = n;
    Modify(s, a[n], -1);
    Modify(s2, a[n], 1);
    while (1) {
      while (sum > k && he) {
        sum -= he - getsum(s, a[he]);
        sum -= getsum(s2, a[he] - 1);
        Modify(s, a[he], -1);
        he--;
      }
      if (!he) break;
      ans += he;
      ta--;
      sum += he - getsum(s, a[ta]);
      sum += getsum(s2, a[ta] - 1);
      Modify(s2, a[ta], 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
