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 = 1e5 + 10;
int fen[2][MAXN];
long long cur, k, ans;
int n, a[MAXN];
map<int, int> mp;
vector<int> R;
void add(int i, bool p, int val) {
  for (i++; i > 0 && i < MAXN;) {
    fen[p][i] += val;
    if (!p)
      i -= i & (-i);
    else
      i += i & (-i);
  }
}
int get(int i, bool p) {
  int res = 0;
  for (i++; i > 0 && i < MAXN;) {
    res += fen[p][i];
    if (!p)
      i += i & (-i);
    else
      i -= i & (-i);
  }
  return res;
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> a[i], R.push_back(a[i]);
  sort(R.begin(), R.end());
  R.resize(unique(R.begin(), R.end()) - R.begin());
  for (int i = 0; i < R.size(); i++) mp[R[i]] = i;
  for (int i = 0; i < n; i++) a[i] = mp[a[i]];
  int j = n - 1;
  while (j >= 0) {
    cur += get(a[j] + 1, 0) + get(a[j] - 1, 1);
    add(a[j], 1, 1);
    j--;
  }
  j++;
  for (int i = 0; i < n; i++) {
    if (i == j) {
      add(a[j], 1, -1);
      cur -= get(a[j] + 1, 0) + get(a[j] - 1, 1);
      j++;
    }
    cur += get(a[i] + 1, 0) + get(a[i] - 1, 1);
    add(a[i], 0, 1);
    while (cur > k && j < n) {
      add(a[j], 1, -1);
      cur -= get(a[j] + 1, 0) + get(a[j] - 1, 1);
      j++;
    }
    ans += n - j;
  }
  cout << ans;
}
