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 = 100010;
int n;
long long k;
int s[100010], inv[100010];
void modify(int x, int d) {
  for (x++; x <= n + 2; x += -x & x) {
    s[x] += d;
  }
}
int get(int x) {
  int res = 0;
  for (x++; x > 0; x &= x - 1) {
    res += s[x];
  }
  return res;
}
int getsum(int L, int R) {
  if (L > R) return 0;
  int res = get(R) - get(L - 1);
  return res;
}
vector<int> compress(vector<int>& a) {
  vector<pair<int, int> > ev(a.size());
  for (int i = 0; i < (int)a.size(); i++) ev[i] = make_pair(a[i], i);
  sort(ev.begin(), ev.end());
  vector<int> real;
  for (int i = 0; i < (int)ev.size(); i++) {
    if (i == 0 || ev[i].first != ev[i - 1].first) real.push_back(ev[i].first);
    a[ev[i].second] = real.size() - 1;
  }
  return real;
}
int main() {
  cin >> n >> k;
  vector<int> a(n);
  for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
  compress(a);
  for (int i = 0; i < (int)(n); i++) {
    inv[i] += getsum(a[i] + 1, n);
    modify(a[i], +1);
  }
  memset(s, 0, sizeof(s));
  for (int i = n - 1; i >= 0; i--) {
    inv[i] += getsum(0, a[i] - 1);
    modify(a[i], +1);
  }
  long long total = 0;
  for (int i = 0; i < (int)(n); i++) total += inv[i];
  total >>= 1;
  long long ans = 0;
  int R = 0;
  memset(s, 0, sizeof(s));
  for (int L = 0; L < (int)(n); L++) {
    if (R < L) R = L;
    while (R < n && (total > k)) {
      total -= inv[R];
      total += getsum(a[R] + 1, n);
      modify(a[R], +1);
      R++;
    }
    if (L) ans += n - R;
    if (R > L) {
      total += inv[L];
      total -= getsum(0, a[L] - 1);
    }
    modify(a[L], -1);
  }
  cout << ans << endl;
  return 0;
}
