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 a[100055], b[100055], bit1[100055], bit2[100055];
map<int, int> mymap;
int bitn, n;
long long int k;
void preparemap() {
  sort(b, b + n);
  bitn = 1;
  mymap[b[0]] = bitn++;
  for (int i = 1; i < n; i++)
    if (b[i] != b[i - 1]) mymap[b[i]] = bitn++;
}
int query(int x, int ar[]) {
  int s = 0;
  while (x > 0) {
    s += ar[x];
    x -= x & (-x);
  }
  return s;
}
void update(int x, int ar[], int inc) {
  while (x <= bitn) {
    ar[x] += inc;
    x += x & (-x);
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  preparemap();
  memset(bit1, 0, sizeof bit1);
  memset(bit2, 0, sizeof bit2);
  long long int ninv = 0;
  long long int ans = 0;
  int p1 = 0, p2;
  for (p2 = n - 1; p2 > 0; p2--) {
    int idx = mymap[a[p2]];
    update(idx, bit2, 1);
    ninv += query(idx - 1, bit2);
  }
  p2 = 1;
  for (p1 = 0; p1 < n; p1++) {
    if (p2 == p1)
      update(mymap[a[p2++]], bit2, -1);
    else {
      int idx = mymap[a[p1]];
      ninv += query(idx - 1, bit2);
      ninv += (p1 - query(idx, bit1));
    }
    update(mymap[a[p1]], bit1, 1);
    while (ninv > k && p2 < n) {
      int idx = mymap[a[p2]];
      ninv -= query(idx - 1, bit2);
      ninv -= (p1 + 1 - query(idx, bit1));
      update(idx, bit2, -1);
      p2++;
    }
    ans += (n - p2);
  }
  cout << ans << endl;
  return 0;
}
