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 stmp;
const int MAX = 100000;
const int INF = 1000000001;
class tree {
 public:
  void init(int n) {
    for (size = 1; size < n; size *= 2)
      ;
    val.assign(2 * size, vector<int>());
  }
  void set(int pos, int v) { val[pos + size].push_back(v); }
  void build() {
    for (int i = (size - 1); i >= (1); --i) {
      val[i].resize(val[2 * i].size() + val[2 * i + 1].size());
      merge((val[2 * i]).begin(), (val[2 * i]).end(), (val[2 * i + 1]).begin(),
            (val[2 * i + 1]).end(), val[i].begin());
    }
  }
  int lower(int b, int e, int v) {
    if (b == e) return 0;
    return query(b + size, e + size, v);
  }
  int greater(int b, int e, int v) { return (e - b) - lower(b, e, v + 1); }

 private:
  int size;
  vector<vector<int> > val;
  int sth(int p, int v) {
    return lower_bound((val[p]).begin(), (val[p]).end(), v) - val[p].begin();
  }
  int query(int p, int q, int v) {
    if (p + 1 == q) return sth(p, v);
    if (p & 1) return sth(p, v) + query(p + 1, q, v);
    if (q & 1) return sth(q - 1, v) + query(p, q - 1, v);
    return query(p / 2, q / 2, v);
  }
};
int tab[MAX];
tree T;
int n;
long long k;
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(0);
  cin >> n >> k;
  T.init(n);
  for (int i = 0; i < (n); ++i) {
    cin >> tab[i];
    T.set(i, tab[i]);
  }
  T.build();
  long long total = 0;
  for (int i = 0; i < (n); ++i) total += T.greater(0, i, tab[i]);
  if (total <= k) {
    cout << (long long)n * (n - 1) / 2 << endl;
    return 0;
  }
  long long res = 0;
  long long inv = 0;
  int j = 1;
  for (int i = (1); i <= (n - 1); ++i) {
    while (j < n && total - inv > k) {
      inv += T.greater(0, i, tab[j]) + T.lower(j + 1, n, tab[j]);
      ++j;
    }
    if (total - inv > k || j == n) break;
    res += (n - j);
    inv -= T.greater(0, i, tab[i]);
    inv -= T.lower(j, n, tab[i]);
  }
  cout << res << endl;
  return 0;
}
