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 mxN = 1e6 + 5, mod = 1e9 + 7;
int n, k, a[mxN], f[mxN], p[mxN], s[mxN];
bool del[mxN];
long long inv;
void comp() {
  vector<int> vec;
  for (int i = 1; i <= n; ++i) vec.push_back(a[i]);
  sort(vec.begin(), vec.end());
  for (int i = 1; i <= n; ++i)
    a[i] = lower_bound(vec.begin(), vec.end(), a[i]) - vec.begin();
}
void Upd(int pos, int v) {
  for (int i = pos; i < n; i |= i + 1) f[i] += v;
}
int Get(int pos) {
  int ans = 0;
  for (int i = pos; i >= 0; i &= i + 1, --i) ans += f[i];
  return ans;
}
void calc() {
  for (int i = 1; i <= n; ++i) {
    p[i] = Get(n - 1) - Get(a[i]);
    inv += p[i];
    Upd(a[i], 1);
  }
  for (int i = 0; i < n; ++i) f[i] = 0;
  for (int i = n; i >= 1; i--) {
    s[i] = Get(a[i] - 1);
    Upd(a[i], 1);
  }
}
int main() {
  ios ::sync_with_stdio(false), cin.tie(nullptr);
  cin >> n >> k;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  comp();
  calc();
  long long ans = (n - 1) * (inv <= k), r = 0;
  for (int i = 0; i < n; ++i) f[i] = 0;
  for (int i = 2; i < n; ++i) {
    if (r < i) {
      inv -= p[i] + s[i];
      r = i;
      Upd(a[i], 1);
    }
    while (r + 1 <= n && inv > k) {
      ++r;
      inv -= p[r] + s[r];
      inv += Get(n - 1) - Get(a[r]);
      Upd(a[r], 1);
    }
    ans += n - r;
    inv += p[i] + s[i];
    inv -= Get(a[i] - 1);
    Upd(a[i], -1);
  }
  cout << ans;
  return 0;
}
