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 N = 110000;
int n, x[N], aib1[N], aib2[N], vmax;
long long rez, k, inv;
pair<int, int> t[N];
void add(int aib[], int poz, int val) {
  for (; poz <= n; poz += poz & (-poz)) aib[poz] += val;
}
inline int sum(int aib[], int poz) {
  int r = 0;
  for (; poz; poz -= poz & (-poz)) r += aib[poz];
  return r;
}
int q(int aib[], int pozx, int pozy) {
  return sum(aib, pozy) - sum(aib, pozx - 1);
}
int main() {
  int i, r, l;
  cin >> n >> k;
  for (i = 1; i <= n; ++i) cin >> x[i], t[i].first = x[i], t[i].second = i;
  sort(t + 1, t + n + 1);
  for (i = 1; i <= n; ++i)
    x[t[i].second] = (t[i - 1].first == t[i].first) ? x[t[i - 1].second]
                                                    : x[t[i - 1].second] + 1,
    vmax = max(vmax, x[t[i].second]);
  r = n - 1;
  for (i = 1; i <= r; ++i) {
    add(aib1, x[i], 1);
    inv = inv + q(aib1, x[i] + 1, vmax);
  }
  for (l = n; l > 1; --l) {
    inv = inv + q(aib1, x[l] + 1, vmax) + q(aib2, 1, x[l] - 1);
    add(aib2, x[l], 1);
    while ((r && inv > k) || r >= l) {
      add(aib1, x[r], -1);
      inv = inv - q(aib1, x[r] + 1, vmax) - q(aib2, 1, x[r] - 1);
      --r;
    }
    if (!r) break;
    rez += r;
  }
  cout << rez;
  return 0;
}
