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 = (1 << 18);
map<int, int> to_c, from_c;
struct seg_tree {
  int t[N * 2 + 228];
  seg_tree(const vector<int> &a) {
    fill(t, t + N * 2 + 228, 0);
    int n = a.size();
    for (int i = 0; i < n; ++i) {
      t[i + N] = a[i];
    }
    for (int i = N - 1; i >= 0; --i) {
      t[i] = t[2 * i] + t[2 * i + 1];
    }
  }
  int get(const int &v, const int &tl, const int &tr, const int &l,
          const int &r) {
    if (l <= tl && tr <= r) {
      return t[v];
    }
    if (l >= tr || r <= tl) {
      return 0;
    }
    int m = (tl + tr) / 2;
    return get(2 * v, tl, m, l, r) + get(2 * v + 1, m, tr, l, r);
  }
  void change(int i, const int &add) {
    i += N;
    t[i] += add;
    while (i != 1) {
      i /= 2;
      t[i] = t[2 * i] + t[2 * i + 1];
    }
  }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cout << fixed << setprecision(15);
  int n;
  long long k;
  cin >> n >> k;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b.begin(), b.end());
  b.resize(unique(b.begin(), b.end()) - b.begin());
  for (int i = 0; i < b.size(); ++i) {
    to_c[b[i]] = i;
    from_c[i] = b[i];
  }
  vector<int> hist(n);
  seg_tree pref(hist), suf(hist);
  int l = 0, r = 1;
  long long cnt = 0;
  long long ans = 0;
  for (int i = n - 1; i >= 0; --i) {
    suf.change(to_c[a[i]], 1);
    cnt += suf.get(1, 0, N, 0, to_c[a[i]]);
  }
  suf.change(to_c[a[0]], -1);
  pref.change(to_c[a[0]], 1);
  for (; r < n; ++r) {
    while (l + 1 < r) {
      long long delta = pref.get(1, 0, N, to_c[a[l + 1]] + 1, n + 1) +
                        suf.get(1, 0, N, 0, to_c[a[l + 1]]);
      if (delta + cnt <= k) {
        pref.change(to_c[a[l + 1]], 1);
        cnt += delta;
        ++l;
      } else {
        break;
      }
    }
    if (cnt <= k) ans += l + 1;
    suf.change(to_c[a[r]], -1);
    cnt -= pref.get(1, 0, N, to_c[a[r]] + 1, n + 1) +
           suf.get(1, 0, N, 0, to_c[a[r]]);
  }
  cout << ans;
  return 0;
}
