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 long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long maxn = 1e5 + 10, maxval = 1e9 + 10, mlg = (1 << 18) + 10;
long long n, k;
long long mp[2][maxn];
long long inv, a[maxn], ans;
void upd(long long ind, long long x, long long tp) {
  for (ind; ind < maxn; ind += ind & (-ind)) {
    mp[tp][ind] += x;
  }
}
long long get(long long ind, long long tp) {
  long long sum = 0;
  for (ind; ind > 0; ind -= ind & (-ind)) {
    sum += mp[tp][ind];
  }
  return sum;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  vector<long long> ve;
  for (int i = 0; i < n; i++) {
    cin >> a[i + 1];
    ve.push_back(a[i + 1]);
  }
  sort(ve.begin(), ve.end());
  ve.resize(unique(ve.begin(), ve.end()) - ve.begin());
  for (int i = 1; i <= n; i++) {
    a[i] = lower_bound(ve.begin(), ve.end(), a[i]) - ve.begin() + 1;
  }
  for (int i = 0; i < n; i++) {
    inv += get(maxn - 1, 1) - get(a[i + 1], 1);
    upd(a[i + 1], 1, 1);
  }
  long long r = 2;
  upd(a[1], -1, 1);
  inv -= get(a[1] - 1, 1);
  for (int l = 1; l <= n; l++) {
    if (r == l) {
      inv -= get(a[r] - 1, 1);
      inv -= get(maxn - 1, 0) - get(a[r], 0);
      upd(a[r], -1, 1);
      r++;
    }
    inv += get(maxn - 1, 0) - get(a[l], 0);
    inv += get(a[l] - 1, 1);
    upd(a[l], 1, 0);
    while ((inv > k) && (r <= n)) {
      inv -= get(a[r] - 1, 1);
      inv -= get(maxn - 1, 0) - get(a[r], 0);
      upd(a[r], -1, 1);
      r++;
    }
    ans += (n + 1 - r);
  }
  cout << ans << endl;
}
