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;
vector<long long> st[4 * 100005];
int n, ar[100005];
long long k;
vector<long long> combine(vector<long long> &res, vector<long long> &a,
                          vector<long long> &b) {
  int i = 0, j = 0;
  while (i < a.size() && j < b.size()) {
    if (a[i] < b[j])
      res.push_back(a[i++]);
    else
      res.push_back(b[j++]);
  }
  while (i < a.size()) {
    res.push_back(a[i++]);
  }
  while (j < b.size()) {
    res.push_back(b[j++]);
  }
  return res;
}
void build(int v, int l, int r) {
  if (l == r) {
    st[v] = {ar[l]};
  } else {
    int md = (l + r) >> 1;
    build(v << 1, l, md);
    build(v << 1 | 1, md + 1, r);
    combine(st[v], st[v << 1], st[v << 1 | 1]);
  }
}
long long query(int v, int l, int r, int ql, int qr, int val,
                bool inverse = false) {
  if (l >= ql && r <= qr) {
    if (!inverse)
      return upper_bound(st[v].begin(), st[v].end(), val - 1) - st[v].begin();
    else
      return st[v].end() - upper_bound(st[v].begin(), st[v].end(), val);
  }
  if (l > qr || r < ql) return 0;
  int md = (l + r) >> 1;
  return query(v * 2, l, md, ql, qr, val, inverse) +
         query(v * 2 + 1, md + 1, r, ql, qr, val, inverse);
}
int main() {
  scanf("%d %I64d", &n, &k);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", ar + i);
  }
  build(1, 1, n);
  long long inversions = 0, ans = 0;
  for (int i = 1; i < n; ++i) {
    inversions += query(1, 1, n, i, n, ar[i]);
  }
  int i = 1, j = 2;
  while (j <= n) {
    if (inversions <= k) {
      ans += (n - j + 1);
      if (i + 1 == j) {
        i++;
        j++;
      } else {
        i++;
        inversions += query(1, 1, n, j, n, ar[i]);
        inversions += query(1, 1, n, 1, i, ar[i], true);
      }
    } else {
      inversions -= query(1, 1, n, j + 1, n, ar[j]);
      inversions -= query(1, 1, n, 1, i, ar[j], true);
      j++;
    }
  }
  cout << ans;
  return 0;
}
