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 nmax = 1e5 + 42;
int n;
long long k;
int arr[nmax];
void compress() {
  set<int> seen = {};
  for (int i = 1; i <= n; i++) seen.insert(arr[i]);
  map<int, int> c = {};
  int val = 0;
  for (auto k : seen) {
    val++;
    c[k] = val;
  }
  for (int i = 1; i <= n; i++) arr[i] = c[arr[i]];
}
int fenwick[nmax][2];
int sum(int pos, int index) {
  int ret = 0;
  while (pos) {
    ret = ret + fenwick[pos][index];
    pos = pos - (pos & (-pos));
  }
  return ret;
}
int query(int l, int r, int index) { return sum(r, index) - sum(l - 1, index); }
void update(int pos, int add, int index) {
  while (pos <= n) {
    fenwick[pos][index] += add;
    pos = pos + (pos & (-pos));
  }
}
int main() {
  scanf("%i%lld", &n, &k);
  for (int i = 1; i <= n; i++) scanf("%i", &arr[i]);
  compress();
  int r = n + 1;
  long long now = 0;
  while (r > 1 && now <= k) {
    r--;
    now = now + query(1, arr[r] - 1, 1);
    if (now > k) {
      now = now - query(1, arr[r] - 1, 1);
      r++;
      break;
    }
    update(arr[r], 1, 1);
  }
  long long ans = 0;
  for (int l = 1; l < n; l++) {
    now = now + query(arr[l] + 1, n, 0) + query(1, arr[l] - 1, 1);
    update(arr[l], 1, 0);
    while (l >= r || (now > k && r <= n)) {
      now = now - query(arr[r] + 1, n, 0) - query(1, arr[r] - 1, 1);
      update(arr[r], -1, 1);
      r++;
    }
    ans = ans + (n - r + 1);
  }
  printf("%lld\n", ans);
  return 0;
}
