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 + 100;
const int mod = 1e9 + 7;
const long long inf = (1ll << 62);
const int base = 31;
long long n, k, a[maxn], lef[maxn], rig[maxn], seg[4 * maxn];
pair<long long, long long> b[maxn];
long long get(int id, int L, int R, int l, int r) {
  if (L == l and R == r) return seg[id];
  long long ret = 0;
  int mid = (L + R) / 2;
  if (mid > l) ret += get(2 * id + 0, L, mid, l, min(mid, r));
  if (mid < r) ret += get(2 * id + 1, mid, R, max(l, mid), r);
  return ret;
}
void change(int id, int L, int R, int idx, int x) {
  seg[id] += x;
  if (R - L == 1) return;
  int mid = (L + R) / 2;
  if (mid > idx)
    change(2 * id + 0, L, mid, idx, x);
  else
    change(2 * id + 1, mid, R, idx, x);
}
long long find_inversion() {
  for (int i = 1; i <= n; i++) {
    lef[i] = get(1, 1, maxn, a[i] + 1, maxn);
    change(1, 1, maxn, a[i], 1);
  }
  memset(seg, 0, sizeof seg);
  long long tmp = 0;
  for (int i = n; i >= 1; i--) {
    rig[i] = get(1, 1, maxn, 1, a[i]);
    change(1, 1, maxn, a[i], 1);
    tmp += rig[i];
  }
  memset(seg, 0, sizeof seg);
  return tmp;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    b[i] = make_pair(a[i], i);
  }
  int cnt = 0;
  sort(b + 1, b + n + 1);
  int ret = 1;
  for (int i = 1; i <= n; i++) {
    cnt += (b[i].first > b[i - 1].first);
    a[b[i].second] = cnt;
  }
  long long inv = find_inversion(), myinv = 0, ans = 0;
  int r = 2;
  for (int l = 1; l < n; l++) {
    r = max(r, l + 1);
    while (inv - k > myinv and r <= n) {
      myinv += lef[r] + rig[r] - get(1, 1, maxn, a[r] + 1, maxn);
      change(1, 1, maxn, a[r], 1);
      r++;
    }
    if (r <= n) ans += n - r + 1;
    if (r > l + 1)
      myinv -= lef[l + 1] + rig[l + 1] - get(1, 1, maxn, 1, a[l + 1]);
    change(1, 1, maxn, a[l + 1], -1);
  }
  cout << ans << endl;
}
