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 V = 100000 + 10;
int n, m;
long long a[V], b[V];
struct Bit {
  int c[V], n;
  int total;
  int sum(int pos) {
    int ret = 0;
    while (pos > 0) {
      ret += c[pos];
      pos -= last_bit(pos);
    }
    return ret;
  }
  int bigger_than(int x) { return total - sum(x); }
  int smaller_than(int x) { return sum(x - 1); }
  int last_bit(int x) { return x & -x; }
  void update(int pos, int x) {
    total += x;
    while (pos <= n) {
      c[pos] += x;
      pos += last_bit(pos);
    }
  }
  void add(int x) { update(x, 1); }
  void sub(int x) { update(x, -1); }
  void init(int _n) {
    n = _n;
    clear();
  }
  void clear() {
    total = 0;
    for (int i = 1; i <= n; ++i) c[i] = 0;
  }
};
Bit bit_left, bit_right;
int main() {
  cin >> n >> m;
  for (int i = 0; i < n; ++i) cin >> a[i];
  for (int i = 0; i < n; ++i) b[i] = a[i];
  sort(b, b + n);
  for (int i = 0; i < n; ++i) a[i] = lower_bound(b, b + n, a[i]) - b + 1;
  for (int i = 0; i < n; ++i) assert(a[i] >= 1 && a[i] <= n);
  bit_left.init(n);
  bit_right.init(n);
  long long total = 0;
  for (int i = n - 1; i >= 1; --i) {
    bit_right.add(a[i]);
    total += bit_right.smaller_than(a[i]);
  }
  long long ans = 0;
  int r = 1;
  for (int l = 0; l < n; ++l) {
    bit_left.add(a[l]);
    total += bit_left.bigger_than(a[l]) + bit_right.smaller_than(a[l]);
    while (r < n && (total > m || r <= l)) {
      total -= bit_left.bigger_than(a[r]) + bit_right.smaller_than(a[r]);
      bit_right.sub(a[r]);
      r++;
    }
    ans += n - r;
  }
  cout << ans << endl;
}
