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 = 1e5 + 5;
int n;
int arr[N], fen[3][N];
long long k, ans;
vector<pair<int, int> > v;
void upd(int pos, int val, int which) {
  for (int i = pos; i <= n; i += (i & -i)) {
    fen[which][i] += val;
  }
}
int get(int pos, int which) {
  int res = 0;
  for (int i = pos; i; i -= (i & -i)) {
    res += fen[which][i];
  }
  return res;
}
int main() {
  scanf("%d", &n);
  scanf("%lld", &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
    v.push_back({arr[i], i});
  }
  sort(v.begin(), v.end());
  for (int i = 0; i < n; i++) {
    if (i == 0)
      arr[v[i].second] = 1;
    else {
      if (v[i].first == v[i - 1].first)
        arr[v[i].second] = arr[v[i - 1].second];
      else
        arr[v[i].second] = arr[v[i - 1].second] + 1;
    }
  }
  long long curInv = 0;
  for (int i = n; i >= 1; i--) {
    curInv += get(arr[i] - 1, 2);
    upd(arr[i], 1, 2);
  }
  puts("");
  int r = 2;
  for (int i = 1; i <= 1; i++) upd(arr[i], 1, 0);
  for (int i = 2; i <= n; i++) upd(arr[i], 1, 1);
  for (int i = 1; i <= n; i++) {
    if (i != 1) {
      if (r == i) {
        upd(arr[r], -1, 1);
        upd(arr[i], 1, 0);
        ++r;
      } else {
        int lInv = get(n, 0) - get(arr[i], 0);
        int rInv = get(arr[i] - 1, 1);
        curInv += (lInv + rInv);
        upd(arr[i], 1, 0);
      }
    }
    while (r <= n and curInv > k) {
      int rInv = get(arr[r] - 1, 1);
      int lInv = get(n, 0) - get(arr[r], 0);
      curInv -= (rInv + lInv);
      upd(arr[r], -1, 1);
      ++r;
    }
    if (r == n + 1) break;
    ans += (n - r + 1);
  }
  printf("%lld\n", ans);
}
