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 tree[500000][3];
map<int, int> name;
int ara[100005];
void update(int node, int st, int en, int l, int v, int ver) {
  if (st > l or en < l) return;
  if (st == en) {
    tree[node][ver] += v;
    return;
  }
  int mid = (st + en) / 2;
  update(node * 2, st, mid, l, v, ver);
  update(node * 2 + 1, mid + 1, en, l, v, ver);
  tree[node][ver] = tree[node * 2][ver] + tree[node * 2 + 1][ver];
}
int query(int node, int st, int en, int l, int r, int ver) {
  if (st > r or en < l or l > r) return 0;
  if (st >= l and en <= r) return tree[node][ver];
  int mid = (st + en) >> 1;
  int p = query(node * 2, st, mid, l, r, ver);
  p += query(node * 2 + 1, mid + 1, en, l, r, ver);
  return p;
}
int main() {
  int i, j, k, l, m, n, cnt = 0;
  vector<int> vc;
  long long K;
  scanf("%d", &n);
  cin >> K;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &ara[i]);
    vc.push_back(ara[i]);
  }
  sort(vc.begin(), vc.end());
  for (int i = 0; i < vc.size(); i++) {
    if (i && vc[i] == vc[i - 1]) continue;
    name[vc[i]] = ++cnt;
  }
  for (int i = 1; i <= n; i++) {
    ara[i] = name[ara[i]];
  }
  long long tot = 0, now = 0, gt = 0, p = 2, ans = 0;
  for (int i = n; i >= 1; i--) {
    tot += query(1, 1, cnt, 1, ara[i] - 1, 2);
    update(1, 1, cnt, ara[i], 1, 2);
  }
  update(1, 1, cnt, ara[1], 1, 0);
  update(1, 1, cnt, ara[1], -1, 2);
  now = 0;
  for (int i = 2; i <= n; i++) {
    if (i > 2) {
      now += query(1, 1, cnt, 1, ara[i - 1] - 1, 2);
      now += query(1, 1, cnt, ara[i - 1] + 1, cnt, 0);
      update(1, 1, cnt, ara[i - 1], -1, 2);
      update(1, 1, cnt, ara[i - 1], 1, 1);
    }
    while (p < i) {
      long long d = query(1, 1, cnt, 1, ara[p] - 1, 2);
      d += query(1, 1, cnt, ara[p] + 1, cnt, 0);
      if (tot - (now - d) <= K) {
        now -= d;
        update(1, 1, cnt, ara[p], -1, 1);
        update(1, 1, cnt, ara[p], 1, 0);
        p++;
      } else
        break;
    }
    if ((tot - now) <= K) ans += (p - 1);
  }
  cout << ans << endl;
}
