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 long long MAX_N = 1e5 + 5;
long long n, k, ans;
long long Arr[MAX_N];
long long Brr[MAX_N];
long long fenL[MAX_N], fenR[MAX_N];
long long getL(long long tmp) {
  long long res = 0;
  for (; tmp; tmp -= tmp & -tmp) res += fenL[tmp];
  return res;
}
long long getR(long long tmp) {
  long long res = 0;
  for (; tmp; tmp -= tmp & -tmp) res += fenR[tmp];
  return res;
}
void updL(long long val, long long tmp) {
  for (; tmp < MAX_N; tmp += tmp & -tmp) fenL[tmp] += val;
}
void updR(long long val, long long tmp) {
  for (; tmp < MAX_N; tmp += tmp & -tmp) fenR[tmp] += val;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < n; i++) {
    cin >> Arr[i];
    Brr[i] = Arr[i];
  }
  sort(Brr, Brr + n);
  long long len = unique(Brr, Brr + n) - Brr;
  for (long long i = 0; i < n; i++)
    Arr[i] = 1 + lower_bound(Brr, Brr + len, Arr[i]) - Brr;
  long long l = 0, r = n - 1, sum = 0;
  updR(+1, Arr[n - 1]);
  while (r && sum + getR(Arr[r - 1] - 1) <= k) {
    sum += getR(Arr[r - 1] - 1);
    updR(+1, Arr[r - 1]);
    r--;
  }
  for (long long i = 0; i < n; i++) {
    sum += getR(Arr[i] - 1);
    sum += getL(MAX_N - 1) - getL(Arr[i]);
    updL(+1, Arr[i]);
    if (r == i) {
      sum -= getR(Arr[i] - 1);
      sum -= getL(MAX_N - 1) - getL(Arr[i]);
      updR(-1, Arr[i]);
      r++;
    }
    while (r < n && k < sum) {
      sum -= getR(Arr[r] - 1);
      sum -= getL(MAX_N - 1) - getL(Arr[r]);
      updR(-1, Arr[r]);
      r++;
    }
    ans += n - r;
  }
  cout << ans << "\n";
  return 0;
}
