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 long long MAXN = 1e5 + 7;
long long a[MAXN], n, k, maxval;
unordered_map<long long, long long> ord;
void condense() {
  set<long long> s;
  for (long long i = 0; i < n; ++i) s.insert(a[i]);
  for (long long cnt = 1; !s.empty();
       maxval = max(maxval, cnt), ++cnt, s.erase(s.begin()))
    ord[*s.begin()] = cnt;
  for (long long i = 0; i < n; ++i) a[i] = ord[a[i]];
}
void update(vector<long long>& b, long long idx, long long val) {
  for (long long i = idx; i <= maxval; i += (i & (-i))) b[i] += val;
}
long long getsum(vector<long long>& b, long long idx) {
  long long s = 0;
  for (long long i = idx; i > 0; i -= (i & (-i))) s += b[i];
  return s;
}
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  condense();
  vector<long long> bitl(maxval + 1, 0), bitr(maxval + 1, 0);
  long long invs = 0, res = 0;
  for (long long i = n - 1; i > 0; --i) {
    invs += getsum(bitr, a[i] - 1);
    update(bitr, a[i], 1);
  }
  invs += getsum(bitr, a[0] - 1);
  update(bitl, a[0], 1);
  if (invs <= k) {
    cout << n * (n - 1) / 2;
    return 0;
  }
  for (long long l = 0, r = 1; r < n;) {
    if (invs > k) {
      invs -= getsum(bitr, a[r] - 1);
      invs -= ((l + 1) - getsum(bitl, a[r]));
      update(bitr, a[r], -1);
      ++r;
    } else {
      res += (n - r);
      ++l;
      invs += getsum(bitr, a[l] - 1);
      invs += (l - getsum(bitl, a[l]));
      update(bitl, a[l], 1);
    }
  }
  cout << res;
}
