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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
const int maxn = (int)1e5 + 10;
int a[maxn], b[maxn], f1[maxn], f2[maxn];
int get(int f[], int pos) {
  int res = 0;
  while (pos >= 0) {
    res += f[pos];
    pos = (pos & (pos + 1)) - 1;
  }
  return res;
}
void add(int f[], int pos, int val) {
  while (pos < maxn) {
    f[pos] += val;
    pos = pos | (pos + 1);
  }
}
int main() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    b[i] = a[i];
  }
  sort(b, b + n);
  int m = unique(b, b + n) - b;
  for (int i = 0; i < n; i++) a[i] = lower_bound(b, b + m, a[i]) - b;
  long long res = 0;
  long long cur = 0;
  for (int i = n - 1; i >= 0; i--) {
    cur += get(f1, a[i] - 1);
    add(f1, a[i], 1);
  }
  int j = n - 1;
  for (int i = n - 1; i > 0; i--) {
    cur += get(f2, a[i] - 1);
    cur += j + 1 - get(f1, a[i]);
    add(f2, a[i], 1);
    while (j >= 0 && (cur > k || j >= i)) {
      cur -= j + 1 - get(f1, a[j]);
      cur -= get(f2, a[j] - 1);
      add(f1, a[j], -1);
      j--;
    }
    res += j + 1;
  }
  cout << res << endl;
  return 0;
}
