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;
long long k;
vector<int> num;
vector<long long> bit;
vector<long long> bit2;
void update_bit(vector<long long>& bit, int i, int d) {
  int x = i;
  while (x <= n) {
    bit[x] += d;
    x += (x & (-x));
  }
}
long long get_sum(vector<long long>& bit, int r) {
  int x = r;
  long long ans = 0;
  while (x > 0) {
    ans += bit[x];
    x -= (x & (-x));
  }
  return ans;
}
long long get_range(vector<long long>& bit, int l, int r) {
  return get_sum(bit, r) - get_sum(bit, l - 1);
}
void solve(int ncase) {
  int k;
  cin >> n >> k;
  num.assign(n, 0);
  bit.assign(n + 1, 0);
  bit2.assign(n + 1, 0);
  for (int i = 0; i < n; i++) {
    cin >> num[i];
  }
  vector<int> sortnum = num;
  sort(sortnum.begin(), sortnum.end());
  sortnum.erase(unique(sortnum.begin(), sortnum.end()), sortnum.end());
  for (int i = 0; i < n; i++) {
    int& a = num[i];
    a = lower_bound(sortnum.begin(), sortnum.end(), a) - sortnum.begin() + 1;
  }
  long long ans = 0;
  long long inv = 0;
  for (int a : num) {
    inv += get_range(bit, a + 1, n);
    update_bit(bit, a, 1);
  }
  int head = 0;
  int l = -1, r = 0;
  while (inv > k) {
    update_bit(bit, num[r], -1);
    inv -= get_range(bit, 1, num[r++] - 1);
  }
  for (int i = r; i < n; i++) {
    while (l < i) {
      long long key = num[l + 1];
      long long tmpinv =
          get_range(bit2, key + 1, n) + get_range(bit, 1, key - 1);
      if (tmpinv + inv <= k) {
        inv += tmpinv;
        l++;
        update_bit(bit2, key, 1);
      } else {
        break;
      }
    }
    if (inv <= k) {
      ans += min(l, i - 1) + 1;
    }
    inv -= get_range(bit2, num[i] + 1, n) + get_range(bit, 1, num[i] - 1);
    update_bit(bit, num[i], -1);
  }
  cout << ans << endl;
}
int main() {
  int T = 1;
  int ncase = 0;
  while (T--) {
    solve(++ncase);
  }
}
