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 N = 1e5 + 5;
long long n, k, a[N], cnt, ans;
vector<long long> vec;
struct fenwick {
  int fen[N];
  void add(int x, int val) {
    for (; x < N; x += x & -x) fen[x] += val;
  }
  int get(int x) {
    int res = 0;
    for (x--; x; x -= x & -x) res += fen[x];
    return res;
  }
} L, R;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vec.push_back(a[i]);
  }
  sort(vec.begin(), vec.end());
  vec.resize(unique(vec.begin(), vec.end()) - vec.begin());
  for (int i = 0; i < n; i++)
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin() + 1;
  for (int i = n - 1; ~i; i--) {
    cnt += R.get(a[i]);
    R.add(a[i], 1);
  }
  if (cnt <= k) return cout << n * (n - 1) / 2 << endl, 0;
  R.add(a[0], -1);
  L.add(n - a[0] + 1, 1);
  for (int e = 2, s = 0; e < n; e++) {
    R.add(a[e - 1], -1);
    cnt -= R.get(a[e - 1]);
    cnt -= L.get(n - a[e - 1] + 1);
    while (cnt <= k) {
      cnt += L.get(n - a[++s] + 1);
      L.add(n - a[s] + 1, 1);
      cnt += R.get(a[s]);
    }
    ans += s;
  }
  cout << ans << endl;
}
