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[maxn], fen2[maxn], a[maxn], num[maxn];
void add(int p, int val) {
  for (p++; p < maxn; p += p & -p) fen[p] += val;
}
int get(int p) {
  int ret = 0;
  for (; p; p -= p & -p) ret += fen[p];
  return ret;
}
void add2(int p, int val) {
  for (p++; p < maxn; p += p & -p) fen2[p] += val;
}
int get2(int p) {
  int ret = 0;
  for (; p; p -= p & -p) ret += fen2[p];
  return ret;
}
int main() {
  int n;
  long long k;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    num[i] = a[i];
  }
  sort(num, num + n);
  for (int i = 0; i < n; i++) a[i] = lower_bound(num, num + n, a[i]) - num + 1;
  long long ans = 0, jav = 0;
  for (int i = n - 1; i >= 0; --i) {
    ans += 1LL * get2(a[i]);
    add2(a[i], 1);
  }
  if (ans <= k) {
    cout << 1LL * n * (n - 1) / 2;
    return 0;
  }
  add2(a[0], -1), add(a[0], 1);
  int cur = 1;
  for (int i = 0; i < n - 1; i++) {
    while (ans > k) {
      ans -= get2(a[cur]);
      ans -= (get(maxn) - get(a[cur] + 1));
      add2(a[cur], -1);
      ++cur;
    }
    if (cur > n) {
      cout << jav << endl;
      return 0;
    }
    jav += 1LL * (n - cur);
    ans += get(maxn) - get(a[i + 1] + 1);
    ans += get2(a[i + 1]);
    add(a[i + 1], 1);
  }
  cout << jav << endl;
}
