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;
struct _ {
  _() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
  }
} _;
const long long base = 311, mod = 1e9 + 7, N = 3e5 + 11, inf = INT_MAX;
long long n, k, r = 1, mx = 0, ans = 0, inverse_count = 0, a[N], tree_left[N],
                tree_right[N];
map<long long, long long> m;
void update_left(long long k, long long x) {
  while (k <= mx) tree_left[k] += x, k += k & -k;
}
void update_right(long long k, long long x) {
  while (k <= mx) tree_right[k] += x, k += k & -k;
}
long long sum_left(long long k) {
  long long sum = 0;
  while (k) sum += tree_left[k], k -= k & -k;
  return sum;
}
long long sum_right(long long k) {
  long long sum = 0;
  while (k) sum += tree_right[k], k -= k & -k;
  return sum;
}
signed main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    m[a[i]] = 1;
  }
  for (auto &x : m) {
    x.second = ++mx;
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = m[a[i]];
  }
  for (long long i = n; i >= 1; i--) {
    inverse_count += sum_right(a[i] - 1);
    update_right(a[i], 1);
  }
  for (long long l = 1; l <= n; l++) {
    inverse_count += (sum_left(mx) - sum_left(a[l])) + sum_right(a[l] - 1);
    update_left(a[l], 1);
    while ((r <= l && r <= n) || (inverse_count > k && r <= n)) {
      inverse_count -= (sum_left(mx) - sum_left(a[r])) + sum_right(a[r] - 1);
      update_right(a[r], -1);
      r++;
    }
    ans += n - r + 1;
  }
  cout << ans;
}
