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;
int n, a[100010], m;
long long k;
map<long long, int> mp;
int st[4 * 100010], CNT[100010], inv[100010];
void update(int idx, int val, int l = 0, int r = m - 1, int p = 1) {
  if (l == r) {
    st[p] += val;
    CNT[l] += val;
    return;
  }
  int mid = (l + r) >> 1;
  if (idx <= mid)
    update(idx, val, l, mid, 2 * p);
  else
    update(idx, val, mid + 1, r, 2 * p + 1);
  st[p] = st[2 * p] + st[2 * p + 1];
}
int q(int a, int b, int l = 0, int r = m - 1, int p = 1) {
  if (a > b) return 0;
  if (b < l || r < a) return 0;
  if (a <= l && r <= b) return st[p];
  int mid = (l + r) >> 1;
  return q(a, b, l, mid, 2 * p) + q(a, b, mid + 1, r, 2 * p + 1);
}
void clearST() {
  for (int i = 0; i < m; i++) update(i, -CNT[i]);
}
int main() {
  scanf("%d%I64d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", a + i), mp[a[i]];
  int cnt = 0;
  for (auto x : mp) mp[x.first] = cnt++;
  m = cnt;
  for (int i = 0; i < n; i++) a[i] = mp[a[i]];
  for (int i = 0; i < n; i++) inv[i] += q(a[i] + 1, m - 1), update(a[i], 1);
  clearST();
  for (int i = n - 1; i >= 0; i--) inv[i] += q(0, a[i] - 1), update(a[i], 1);
  clearST();
  long long totalInv = 0, ans = 0;
  for (int i = 0; i < n; i++) totalInv += 1LL * inv[i];
  totalInv /= 2;
  int r = 1;
  for (int l = 0; l + 1 < n; l++) {
    while (totalInv > k && r < n - 1)
      totalInv += -1LL * inv[r] + 1LL * q(a[r] + 1, m - 1), update(a[r], 1),
          r++;
    if (totalInv <= k) ans += 1LL * (n - r);
    if (l + 1 == r)
      r++;
    else
      totalInv += 1LL * inv[l + 1], update(a[l + 1], -1),
          totalInv -= 1LL * q(0, a[l + 1] - 1);
  }
  printf("%I64d\n", ans);
  return 0;
}
