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 MAXN = 1e5 + 6;
set<long long> as;
map<long long, long long> newa;
long long fenl[MAXN], fenr[MAXN], a[MAXN], n, k;
void updatel(int idx, int val) {
  idx++;
  for (; idx < MAXN; idx += idx & (-idx)) {
    fenl[idx] += val;
  }
}
void updater(int idx, int val) {
  idx++;
  for (; idx < MAXN; idx += idx & (-idx)) {
    fenr[idx] += val;
  }
}
long long queryl(int idx) {
  idx++;
  long long ret = 0;
  for (; idx > 0; idx -= idx & (-idx)) {
    ret += fenl[idx];
  }
  return ret;
}
long long queryr(int idx) {
  long long ret = 0;
  idx++;
  for (; idx > 0; idx -= idx & (-idx)) {
    ret += fenr[idx];
  }
  return ret;
}
int main() {
  cin >> n >> k;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    as.insert(a[i]);
  }
  int n1 = 0;
  for (auto t : as) {
    newa[t] = ++n1;
  }
  for (long long i = 1; i <= n; i++) {
    a[i] = newa[a[i]];
  }
  long long cur_inv = 0;
  for (long long i = n; i >= 1; i--) {
    cur_inv += queryr(a[i] - 1);
    updater(a[i], 1);
  }
  long long r = 1, ans = 0;
  for (long long l = 1; l < n; l++) {
    if (r == l) {
      cur_inv -= (l - 1) - queryl(a[r]) + queryr(a[r] - 1);
      updater(a[r], -1);
      r++;
    }
    cur_inv += (l - 1) - queryl(a[l]);
    cur_inv += queryr(a[l] - 1);
    updatel(a[l], +1);
    while (cur_inv > k && r <= n) {
      cur_inv -= l - queryl(a[r]) + queryr(a[r] - 1);
      updater(a[r], -1);
      r++;
    }
    ans += (n - r + 1);
  }
  cout << ans << endl;
  return 0;
}
