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[100000 + 5];
vector<int> vec;
struct treearr {
  int a[100000 + 5];
  void add(int x, int v) {
    for (; x <= 100000; x += x & -x) a[x] += v;
  }
  int ask(int x) {
    int res = 0;
    for (; x; x -= x & -x) res += a[x];
    return res;
  }
} tb, te;
int main() {
  int n;
  long long k;
  scanf("%d%I64d", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) vec.push_back(a[i]);
  sort(vec.begin(), vec.end());
  vec.erase(unique(vec.begin(), vec.end()), vec.end());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin() + 1;
  long long sum = 0;
  for (int i = n; i >= 2; i--) sum += te.ask(a[i] - 1), te.add(a[i], 1);
  sum += te.ask(a[1] - 1), tb.add(a[1], 1);
  if (sum <= k)
    printf("%I64d\n", 1ll * n * (n - 1) / 2);
  else {
    int ptr = 2;
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      while (ptr <= n && sum > k) {
        sum -= i - tb.ask(a[ptr]) + te.ask(a[ptr] - 1);
        te.add(a[ptr++], -1);
      }
      ans += n - ptr + 1;
      if (i < n)
        sum += i - tb.ask(a[i + 1]) + te.ask(a[i + 1] - 1), tb.add(a[i + 1], 1);
    }
    printf("%I64d\n", ans);
  }
  return 0;
}
