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 = 1e5 + 500;
int n, A[N];
long long k, cur;
vector<int> saz;
int loga[N][2], siz[2];
void add(int x, int y, int k) {
  siz[k] += y;
  for (x += 10; x < N; x += x & -x) loga[x][k] += y;
}
int query(int x, int k) {
  int ret = 0;
  for (x += 10; x; x -= x & -x) ret += loga[x][k];
  return ret;
}
inline int rub(int x) { return query(x - 1, 1) + siz[0] - query(x, 0); }
int main() {
  scanf("%d%lld", &n, &k);
  for (int i = 0; i < n; i++) {
    scanf("%d", A + i);
    saz.push_back(A[i]);
  }
  sort(saz.begin(), saz.end());
  saz.erase(unique(saz.begin(), saz.end()), saz.end());
  for (int i = 0; i < n; i++)
    A[i] = lower_bound(saz.begin(), saz.end(), A[i]) - saz.begin();
  for (int i = 0; i < n; i++) {
    cur += siz[1] - query(A[i], 1);
    add(A[i], 1, 1);
  }
  int l = 0, r = 0;
  long long ans = 0;
  for (; l < n; l++) {
    if (l == r) {
      cur -= rub(A[r]);
      add(A[r], -1, 1);
      r++;
    }
    cur += rub(A[l]);
    add(A[l], 1, 0);
    while (r < n && cur > k) {
      cur -= rub(A[r]);
      add(A[r], -1, 1);
      r++;
    }
    ans += n - r;
  }
  printf("%lld\n", ans);
  return 0;
}
