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, i, ltree[400005], rtree[400005], a[100005], lx, ly, lval;
long long k;
void update(int l, int r, int ind, int* tree, int val) {
  if (l == r) {
    tree[ind] += val;
    return;
  }
  int mid = (l + r) / 2;
  if (lx <= mid)
    update(l, mid, ind * 2, tree, val);
  else
    update(mid + 1, r, ind * 2 + 1, tree, val);
  tree[ind] = tree[ind * 2] + tree[ind * 2 + 1];
}
void query(int l, int r, int ind, int* tree) {
  if (lx <= l && r <= ly) {
    lval += tree[ind];
    return;
  }
  int mid = (l + r) / 2;
  if (lx <= mid) query(l, mid, ind * 2, tree);
  if (ly > mid) query(mid + 1, r, ind * 2 + 1, tree);
}
int main() {
  scanf("%d%lld", &n, &k);
  vector<int> tmp;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    tmp.push_back(a[i]);
  }
  sort(tmp.begin(), tmp.end());
  tmp.resize(unique(tmp.begin(), tmp.end()) - tmp.begin());
  for (i = 0; i < n; i++)
    a[i] = lower_bound(tmp.begin(), tmp.end(), a[i]) - tmp.begin() + 1;
  long long tot = 0;
  for (i = 0; i < n; i++) {
    lx = a[i] + 1;
    ly = tmp.size();
    if (lx <= ly) {
      lval = 0;
      query(1, n, 1, ltree);
      tot += lval;
    }
    lx = a[i];
    update(1, n, 1, ltree, 1);
  }
  int p = n - 1;
  long long res = 0;
  for (i = n - 1; i >= 0; i--) {
    lx = 1;
    ly = a[i] - 1;
    if (lx <= ly) {
      lval = 0;
      query(1, n, 1, rtree);
      tot += lval;
    }
    lx = a[i] + 1;
    ly = tmp.size();
    if (lx <= ly) {
      lval = 0;
      query(1, n, 1, ltree);
      tot += lval;
    }
    lx = a[i];
    update(1, n, 1, rtree, 1);
    while (p >= 0 && (p >= i || tot > k)) {
      lx = a[p] + 1;
      ly = tmp.size();
      if (lx <= ly) {
        lval = 0;
        query(1, n, 1, ltree);
        tot -= lval;
      }
      lx = 1;
      ly = a[p] - 1;
      if (lx <= ly) {
        lval = 0;
        query(1, n, 1, rtree);
        tot -= lval;
      }
      lx = a[p];
      update(1, n, 1, ltree, -1);
      p--;
    }
    if (p < i && tot <= k) {
      res += p + 1;
    }
  }
  printf("%lld\n", res);
}
