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 + 5;
int A[MAXN], B[MAXN], fwt1[MAXN] = {0}, fwt2[MAXN] = {0};
int n;
unordered_map<int, int> M;
void update(int* T, int x, int d) {
  for (; x < MAXN; x += x & -x) T[x] += d;
}
long long query(int* T, int x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) sum += T[x];
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) cin >> A[i], B[i] = A[i];
  sort(B, B + n);
  int p = unique(B, B + n) - B;
  for (int i = 0; i < p; i++) M[B[i]] = i + 1;
  for (int i = 0; i < n; i++) B[i] = M[A[i]];
  int l = n - 2;
  long long curr = 0, ans = 0;
  for (int i = 0; i < n - 1; i++) {
    curr += query(fwt1, MAXN) - query(fwt1, B[i]);
    update(fwt1, B[i], 1);
  }
  int z = 0;
  for (int r = n - 1; r >= 0; r--) {
    update(fwt2, B[r], 1);
    curr += query(fwt1, MAXN) - query(fwt1, B[r]);
    curr += query(fwt2, B[r] - 1);
    while (l >= r || (l >= 0 && curr > k)) {
      curr -= query(fwt1, MAXN) - query(fwt1, B[l]);
      curr -= query(fwt2, B[l] - 1);
      update(fwt1, B[l], -1);
      l--;
    }
    ans += l + 1;
  }
  cout << ans << "\n";
}
