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 = 100011;
int pre[maxn << 2];
int sub[maxn << 2];
int n;
int a[maxn];
void PushUp(int* seg, int rt) { seg[rt] = seg[rt << 1] + seg[rt << 1 | 1]; }
void build(int* seg, int l, int r, int rt) {
  if (l == r) {
    seg[l] = 0;
    return;
  }
  int m = (l + r) >> 1;
  build(seg, l, m, rt << 1);
  build(seg, m + 1, r, rt << 1 | 1);
}
void update(int* seg, int p, int c, int l, int r, int rt) {
  if (l == r) {
    seg[rt] += c;
    return;
  }
  int m = (l + r) >> 1;
  if (p <= m)
    update(seg, p, c, l, m, rt << 1);
  else
    update(seg, p, c, m + 1, r, rt << 1 | 1);
  PushUp(seg, rt);
}
int query(int* seg, int L, int R, int l, int r, int rt) {
  if (L <= l && r <= R) {
    return seg[rt];
  }
  int m = (l + r) >> 1;
  int ans = 0;
  if (L <= m) ans += query(seg, L, R, l, m, rt << 1);
  if (R > m) ans += query(seg, L, R, m + 1, r, rt << 1 | 1);
  return ans;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  long long k;
  cin >> n >> k;
  vector<int> v;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back(a[i]);
  }
  sort(v.begin(), v.end());
  v.resize(unique(v.begin(), v.end()) - v.begin());
  for (int i = 1; i <= n; i++)
    a[i] = lower_bound(v.begin(), v.end(), a[i]) - v.begin() + 1;
  int m = v.size() + 1;
  build(pre, 1, m, 1);
  build(sub, 1, m, 1);
  long long cur = 0;
  for (int i = 1; i <= n; i++) {
    cur += (long long)query(pre, a[i] + 1, m, 1, m, 1);
    update(pre, a[i], 1, 1, m, 1);
  }
  long long ans = 0;
  update(pre, a[n], -1, 1, m, 1);
  update(sub, a[n], 1, 1, m, 1);
  for (int l = n - 1, r = n; l >= 1; l--) {
    while (r <= n && cur > k) {
      update(sub, a[r], -1, 1, m, 1);
      cur -= (long long)query(pre, a[r] + 1, m, 1, m, 1);
      if (a[r] > 1) cur -= (long long)query(sub, 1, a[r] - 1, 1, m, 1);
      r++;
    }
    ans += n + 1 - r;
    update(pre, a[l], -1, 1, m, 1);
    cur -= (long long)query(pre, a[l] + 1, m, 1, m, 1);
    if (a[l] > 1) cur -= (long long)query(sub, 1, a[l] - 1, 1, m, 1);
    while (r > l - 1 + 1 && cur <= k) {
      r--;
      update(sub, a[r], 1, 1, m, 1);
      cur += (long long)query(pre, a[r] + 1, m, 1, m, 1);
      if (a[r] > 1) cur += (long long)query(sub, 1, a[r] - 1, 1, m, 1);
    }
  }
  cout << ans << endl;
  return 0;
}
