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 int maxN = 1e5 + 10;
long long n, a[maxN], data[2][4 * maxN], tot[maxN], sz;
long long k;
void add(long long h, long long p, long long x, long long s = 0,
         long long e = sz, long long id = 1) {
  data[h][id] += x;
  if (e - s < 2) return;
  long long mid = s + e >> 1;
  if (p < mid)
    add(h, p, x, s, mid, 2 * id);
  else
    add(h, p, x, mid, e, 2 * id + 1);
}
long long get(long long h, long long l, long long r, long long s = 0,
              long long e = sz, long long id = 1) {
  if (e <= l || r <= s) return 0;
  if (l <= s && e <= r) return data[h][id];
  long long mid = s + e >> 1;
  return get(h, l, r, s, mid, 2 * id) + get(h, l, r, mid, e, 2 * id + 1);
}
int32_t main() {
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    tot[sz++] = a[i];
  }
  sort(tot, tot + sz);
  sz = unique(tot, tot + sz) - tot;
  for (long long i = 0; i < n; i++)
    a[i] = lower_bound(tot, tot + sz, a[i]) - tot;
  add(0, a[0], 1);
  add(1, a[n - 1], 1);
  long long cur = (a[0] > a[n - 1]);
  long long res = 0;
  long long l = 1, r = n - 1;
  for (l; l < n; l++) {
    while (r > l) {
      long long inv = get(1, 0, a[r - 1]) + get(0, a[r - 1] + 1, n);
      if (cur + inv <= k) {
        cur += inv;
        add(1, a[r - 1], 1);
        r--;
      } else
        break;
    }
    while (r < n - 1 && cur > k) {
      cur -= get(1, 0, a[r]) + get(0, a[r] + 1, n);
      add(1, a[r], -1);
      r++;
    }
    if (cur > k) break;
    if (l == r) return cout << (n * (n - 1)) / 2 << endl, 0;
    res += (n - r);
    add(0, a[l], 1);
    cur += get(0, a[l] + 1, n) + get(1, 0, a[l]);
  }
  cout << res << endl;
  return 0;
}
