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 maxm = 1e5 + 5;
struct BIT {
  long long c[maxm];
  long long lowbit(long long i) { return i & -i; }
  void add(long long i, long long t) {
    while (i < maxm) c[i] += t, i += lowbit(i);
  }
  long long ask(long long i) {
    long long ans = 0;
    while (i) ans += c[i], i -= lowbit(i);
    return ans;
  }
} T1, T2;
long long a[maxm];
long long b[maxm];
long long n, k;
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (long long i = 1; i <= n; i++) {
    b[i] = a[i];
  }
  sort(b + 1, b + 1 + n);
  long long num = unique(b + 1, b + 1 + n) - b - 1;
  for (long long i = 1; i <= n; i++) {
    a[i] = lower_bound(b + 1, b + 1 + num, a[i]) - b;
  }
  long long ans = 0;
  long long r = 2;
  long long cnt = 0;
  for (long long i = n; i >= 2; i--) {
    T2.add(a[i], 1);
    cnt += T2.ask(a[i] - 1);
  }
  for (long long l = 1; l <= n; l++) {
    T1.add(a[l], 1);
    cnt += l - T1.ask(a[l]);
    cnt += T2.ask(a[l] - 1);
    while (l == r || (r <= n && cnt > k)) {
      T2.add(a[r], -1);
      cnt -= T2.ask(a[r] - 1);
      cnt -= l - T1.ask(a[r]);
      r++;
    }
    if (cnt > k || r > n) break;
    ans += n - r + 1;
  }
  cout << ans << endl;
  return 0;
}
