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 n, a[100010];
int b[100010], lb;
long long int k;
int big[100010], small[100010];
int lowbit(int x) { return x & -x; }
void addbig(int x, int y) {
  while (x) {
    big[x] += y;
    x -= lowbit(x);
  }
}
int querybig(int x) {
  int res = 0;
  while (x <= lb) {
    res += big[x];
    x += lowbit(x);
  }
  return res;
}
void addsmall(int x, int y) {
  while (x <= lb) {
    small[x] += y;
    x += lowbit(x);
  }
}
int querysmall(int x) {
  int res = 0;
  while (x) {
    res += small[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  lb = unique(b, b + n) - b;
  for (int i = 0; i < n; i++) a[i] = lower_bound(b, b + lb, a[i]) - b + 1;
  long long int w = 0;
  memset(big, 0, sizeof(big));
  for (int i = 0; i < n; i++) {
    w += querybig(a[i] + 1);
    addbig(a[i], 1);
  }
  if (w <= k) {
    printf("%I64d\n", n * (n - 1LL) / 2);
    return 0;
  }
  long long int ans = 0;
  memset(big, 0, sizeof(big));
  memset(small, 0, sizeof(small));
  for (int i = 1; i < n; i++) addsmall(a[i], 1);
  int l = 0, r = 1;
  for (; l < n; l++) {
    if (l > 0) {
      w += querybig(a[l] + 1);
      w += querysmall(a[l] - 1);
    }
    addbig(a[l], 1);
    while (w > k && r < n) {
      w -= querybig(a[r] + 1);
      w -= querysmall(a[r] - 1);
      addsmall(a[r], -1);
      r++;
    }
    if (r >= n) break;
    ans += n - r;
  }
  printf("%I64d\n", ans);
  return 0;
}
