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;
long long n, k;
struct ff {
  int value, order;
} a[100010];
bool cmp(ff a, ff b) { return a.value < b.value; }
long long sum1[100010], sum2[100010];
void add(long long s[], int i, int x) {
  while (i <= n) {
    s[i] += x;
    i += i & -i;
  }
}
long long getsum(long long s[], int i) {
  long long res = 0;
  while (i > 0) {
    res += s[i];
    i -= i & -i;
  }
  return res;
}
int b[100010];
int main() {
  while (~scanf("%I64d%I64d", &n, &k)) {
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i].value);
      a[i].order = i;
      b[i - 1] = a[i].value;
    }
    sort(b, b + n);
    int m = unique(b, b + n) - b;
    for (int i = 1; i <= n; i++) {
      a[i].order = lower_bound(b, b + m, a[i].value) - b + 1;
    }
    memset(sum1, 0, sizeof sum1);
    memset(sum2, 0, sizeof sum2);
    long long res = 0;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      add(sum1, a[i].order, 1);
      res += i - getsum(sum1, a[i].order);
    }
    if (res <= k) {
      printf("%I64d\n", n * (long long)(n - 1) / 2);
      continue;
    }
    add(sum1, a[n].order, -1);
    add(sum2, a[n].order, 1);
    for (int l = n - 1, r = n - 1; r > 0; r--) {
      while (l > 0 && res > k) {
        res -= l - getsum(sum1, a[l].order);
        res -= getsum(sum2, a[l].order - 1);
        add(sum1, a[l].order, -1);
        l--;
      }
      if (l == 0) break;
      ans += l;
      res += getsum(sum2, a[r].order - 1);
      res += l - getsum(sum1, a[r].order);
      add(sum2, a[r].order, 1);
    }
    cout << ans << endl;
  }
  return 0;
}
