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 a[100010], b[100010];
int tree1[(1 << 18) + 10], tree2[(1 << 18) + 10];
class SegTree {
 private:
  int* tree;

 public:
  SegTree(int* tree) : tree(tree) {}
  void insert(int id, int l, int r, int x) {
    if (x > x || x < l || r < x) return;
    if (x <= l && r <= x) {
      ++tree[id];
      return;
    }
    int mid = (l + r) / 2, lid = id * 2, rid = id * 2 + 1;
    insert(lid, l, mid, x);
    insert(rid, mid + 1, r, x);
    tree[id] = tree[lid] + tree[rid];
  }
  void remove(int id, int l, int r, int x) {
    if (x > x || x < l || r < x) return;
    if (x <= l && r <= x) {
      --tree[id];
      return;
    }
    int mid = (l + r) / 2, lid = id * 2, rid = id * 2 + 1;
    remove(lid, l, mid, x);
    remove(rid, mid + 1, r, x);
    tree[id] = tree[lid] + tree[rid];
  }
  int query(int id, int l, int r, int start, int end) {
    if (start > end || end < l || r < start) return 0;
    if (start <= l && r <= end) {
      return tree[id];
    }
    int mid = (l + r) / 2, lid = id * 2, rid = id * 2 + 1;
    return query(lid, l, mid, start, end) + query(rid, mid + 1, r, start, end);
  }
};
int main() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) b[i] = a[i];
  sort(b + 1, b + 1 + n);
  for (int i = 1; i <= n; ++i) {
    int* it = lower_bound(b + 1, b + 1 + n, a[i]);
    a[i] = distance(b, it);
  }
  SegTree t1(tree1), t2(tree2);
  long long inversions = 0;
  for (int i = n; i >= 1; --i) {
    inversions += t2.query(1, 1, n, 1, a[i] - 1);
    if (i > 1) t2.insert(1, 1, n, a[i]);
  }
  long long res = 0;
  t1.insert(1, 1, n, a[1]);
  int l = 1, r = 2;
  for (; l < n; ++l) {
    while (r < n && (inversions > k || l == r)) {
      t2.remove(1, 1, n, a[r]);
      int cnt1 = t1.query(1, 1, n, a[r] + 1, n);
      int cnt2 = t2.query(1, 1, n, 1, a[r] - 1);
      inversions -= cnt1 + cnt2;
      ++r;
    }
    if (l < r && inversions <= k) res += (n - r + 1);
    if (l < n) {
      t1.insert(1, 1, n, a[l + 1]);
      int cnt1 = t1.query(1, 1, n, a[l + 1] + 1, n);
      int cnt2 = t2.query(1, 1, n, 1, a[l + 1] - 1);
      inversions += cnt1 + cnt2;
    }
  }
  cout << res << endl;
  return 0;
}
