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;
long long n, k, a[100005], BITL[100005 + 1], BITR[100005 + 1];
void compress() {
  set<long long> s;
  for (int i = 0; i < n; i++) s.insert(a[i]);
  unordered_map<long long, long long> m;
  long long i = 1;
  for (long long e : s) m[e] = i++;
  for (int i = 0; i < n; i++) a[i] = m[a[i]];
}
void update(long long* BIT, long long id, int val) {
  while (id <= 100005) {
    BIT[id] += val;
    id += (id & -id);
  }
}
long long query(long long* BIT, long long id) {
  long long ans = 0;
  while (id > 0) {
    ans += BIT[id];
    id -= (id & -id);
  }
  return ans;
}
long long solve() {
  long long invs = 0, ans = 0;
  for (int i = 0; i < n - 1; i++) {
    invs += query(BITR, a[n - i - 1] - 1);
    update(BITR, a[n - i - 1], 1);
  }
  long long j = 1;
  for (int i = 0; i < n - 1; i++) {
    if (j == i) {
      invs -= query(BITR, a[j] - 1);
      invs -= query(BITL, 100005) - query(BITL, a[j]);
      update(BITR, a[j++], -1);
    }
    invs += query(BITL, 100005) - query(BITL, a[i]);
    while (j < n && query(BITR, a[i] - 1) + invs > k) {
      invs -= query(BITR, a[j] - 1);
      invs -= query(BITL, 100005) - query(BITL, a[j]);
      update(BITR, a[j++], -1);
    }
    invs += query(BITR, a[i] - 1);
    ans += n - j;
    update(BITL, a[i], 1);
  }
  return ans;
}
int main() {
  cin >> n;
  cin >> k;
  for (int i = 0; i < n; i++) cin >> a[i];
  compress();
  cout << solve() << endl;
  return 0;
}
