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 N = 100005;
int st[N << 2][2], aux[N], vet[N];
void update(int id, int p, int l, int r, int pos, int val) {
  if (l > pos || r < pos) return;
  if (l == r) {
    st[p][id] += val;
    return;
  }
  int mid = (l + r) >> 1;
  update(id, p << 1, l, mid, pos, val);
  update(id, (p << 1) + 1, mid + 1, r, pos, val);
  st[p][id] = st[p << 1][id] + st[(p << 1) + 1][id];
}
int query(int id, int p, int l, int r, int a, int b) {
  if (l > b || r < a) return 0;
  if (l >= a && r <= b) return st[p][id];
  int mid = (l + r) >> 1;
  return query(id, p << 1, l, mid, a, b) +
         query(id, (p << 1) + 1, mid + 1, r, a, b);
}
int main() {
  int n;
  long long k;
  scanf("%d %lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", &aux[i]);
    vet[i] = aux[i];
  }
  sort(aux, aux + n);
  for (int i = 0; i < n; i++) {
    vet[i] = upper_bound(aux, aux + n, vet[i]) - aux;
  }
  int r = 1;
  long long inv_esq = 0LL, inv_dir = 0LL;
  for (int i = n - 1; i >= 1; i--) {
    inv_dir += (query(1, 1, 1, n, 1, vet[i] - 1));
    update(1, 1, 1, n, vet[i], 1);
  }
  long long ans = 0LL, now = 0LL;
  for (int l = 0; l < n; l++) {
    update(0, 1, 1, n, vet[l], 1);
    inv_esq += query(0, 1, 1, n, vet[l] + 1, n);
    now += (query(1, 1, 1, n, 1, vet[l] - 1));
    while (r < n && (r <= l || inv_esq + inv_dir + now > k)) {
      update(1, 1, 1, n, vet[r], -1);
      inv_dir -= (query(1, 1, 1, n, 1, vet[r] - 1));
      now -= query(0, 1, 1, n, vet[r] + 1, n);
      r++;
    }
    if (r < n) ans += (n - r);
  }
  printf("%lld\n", ans);
}
