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;
const int maxn = 100010;
int a[maxn], Tmp[maxn];
int L[maxn], R[maxn];
long long X[maxn];
int n, m, k;
int sum(int pos, int C[]) {
  int ret = 0;
  for (int i = pos; i > 0; i -= i & (-i)) ret += C[i];
  return ret;
}
void add(int pos, int val, int C[]) {
  for (int i = pos; i <= m; i += i & (-i)) C[i] += val;
}
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    Tmp[i] = a[i];
  }
  sort(Tmp + 1, Tmp + 1 + n);
  m = unique(Tmp + 1, Tmp + 1 + n) - Tmp;
  for (int i = 1; i <= n; ++i) {
    a[i] = lower_bound(Tmp + 1, Tmp + m, a[i]) - Tmp;
  }
  for (int i = n; i > 1; --i) {
    X[i] = X[i + 1] + sum(a[i] - 1, R);
    add(a[i], 1, R);
  }
  long long ans = 0, tmp = 0;
  int j = 0;
  for (int i = 2; i <= n; ++i) {
    ++j;
    while (j < i) {
      long long _t = sum(a[j] - 1, R) + sum(m, L) - sum(a[j], L);
      if (_t + tmp + X[i] <= k) {
        add(a[j], 1, L);
        ++j;
        tmp += _t;
      } else
        break;
    }
    --j;
    ans += j;
    tmp -= (sum(m, L) - sum(a[i], L));
    add(a[i], -1, R);
  }
  cout << ans << endl;
  return 0;
}
