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 N = 1e5 + 5;
long long A[N], r, n, k, cnt;
long long ans;
struct fen {
  long long D[N];
  inline fen() { memset(D, 0, sizeof(D)); }
  long long get(long long ind) {
    return (!ind ? 0 : D[ind] + get(ind -= ind & (-ind)));
  }
  void add(long long ind, long long d) {
    for (ind++; ind < N; ind += ind & (-ind)) D[ind] += d;
  }
};
fen fl, fr;
vector<long long> v;
int32_t main() {
  scanf("%I64d %I64d", &n, &k);
  for (long long i = 0; i < n; i++) scanf("%I64d", &A[i]), v.push_back(A[i]);
  set<long long> st(v.begin(), v.end());
  v.assign(st.begin(), st.end());
  for (long long i = 0; i < n; i++)
    A[i] = lower_bound(v.begin(), v.end(), A[i]) - v.begin() + 1;
  for (long long i = n - 1; i >= 0; i--) {
    cnt += fr.get(A[i]);
    fr.add(A[i], 1);
  }
  for (long long i = 0; i < n; i++) {
    cnt += fr.get(A[i]) + i - fl.get(A[i] + 1);
    fl.add(A[i], 1);
    while ((r <= i || cnt > k) && r < n) {
      cnt -= fr.get(A[r]) + i + 1 - fl.get(A[r] + 1);
      fr.add(A[r++], -1);
    }
    ans += n - r;
  }
  printf("%I64d", ans);
}
