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 = 100 * 1000 + 100;
long long sum[2][4 * maxn], a[maxn];
set<long long> s;
map<long long, long long> m;
long long n, k, t = 1, in = 0, ans = 0, c = 0;
void update(long long x, long long l, long long r, long long xx, long long f,
            long long d) {
  if (l > xx || r < xx) return;
  if (l == xx && r == xx) {
    sum[d][x] += f;
    return;
  }
  long long mid = (l + r) / 2;
  update(2 * x, l, mid, xx, f, d);
  update(2 * x + 1, mid + 1, r, xx, f, d);
  sum[d][x] = sum[d][2 * x] + sum[d][2 * x + 1];
}
long long query(long long x, long long l, long long r, long long xx,
                long long yy, long long d) {
  if (l > yy || r < xx) return 0;
  if (l >= xx && r <= yy) return sum[d][x];
  long long mid = (l + r) / 2;
  return query(2 * x, l, mid, xx, yy, d) +
         query(2 * x + 1, mid + 1, r, xx, yy, d);
}
int main() {
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    s.insert(a[i]);
  }
  while (s.size() > 0) {
    m[*s.begin()] = t++;
    s.erase(s.begin());
  }
  for (long long i = 0; i < n; i++) {
    update(1, 0, t, m[a[i]], 1, 1);
    in += query(1, 0, t, m[a[i]] + 1, t, 1);
  }
  for (long long i = 0; i < n - 1; i++) {
    while (c <= i) {
      update(1, 0, t, m[a[c]], -1, 1);
      in -= (query(1, 0, t, 0, m[a[c]] - 1, 1) +
             query(1, 0, t, m[a[c]] + 1, t, 0));
      c++;
    }
    update(1, 0, t, m[a[i]], 1, 0);
    in += query(1, 0, t, m[a[i]] + 1, t, 0) + query(1, 0, t, 0, m[a[i]] - 1, 1);
    while (in > k && c < n) {
      update(1, 0, t, m[a[c]], -1, 1);
      in -= (query(1, 0, t, 0, m[a[c]] - 1, 1) +
             query(1, 0, t, m[a[c]] + 1, t, 0));
      c++;
    }
    if (in <= k) ans += n - c;
  }
  cout << ans << endl;
  return 0;
}
