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 in[101000];
int in2[101000];
int n;
long long k;
map<int, int> xx;
int tre[101000];
int tb[101000];
int xn;
inline void add(int* tr, int x, int c = 1) {
  for (; x <= xn; x += x & -x) tr[x] += c;
}
inline int que(int* tr, int x) {
  int r = 0;
  for (; x; x -= x & -x) r += tr[x];
  return r;
}
inline long long solve() {
  long long ss = 0;
  int i, j;
  for (j = n - 1; j >= 0; j--) {
    int x = que(tb, in[j] - 1);
    if (ss + x > k) {
      j++;
      break;
    }
    ss += x;
    add(tb, in[j]);
  }
  if (j == -1) return n * (n - 1ll) / 2;
  long long ret = 0;
  for (i = 0; i < n; i++) {
    ss += que(tre, xn + 1 - in[i] - 1);
    ss += que(tb, in[i] - 1);
    add(tre, xn + 1 - in[i]);
    for (; j < n; j++) {
      if (ss <= k) break;
      ss -= que(tre, xn + 1 - in[j] - 1);
      ss -= que(tb, in[j] - 1);
      add(tb, in[j], -1);
    }
    if (j == n) break;
    ret += n - j;
  }
  return ret;
}
int main() {
  int i, j;
  scanf("%d%I64d", &n, &k);
  for (i = 0; i < n; i++) {
    scanf("%d", &in[i]);
    in2[i] = in[i];
  }
  sort(in2, in2 + n);
  xn = unique(in2, in2 + n) - in2;
  for (i = 0; i < xn; i++) xx[in2[i]] = i;
  for (i = 0; i < n; i++) in[i] = xx[in[i]] + 1;
  printf("%I64d\n", solve());
}
