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;
long long const MAXN = 1e5 + 8;
long long const INF = 1e12 + 8;
long long const delta = 1000000007;
long long fenl[MAXN];
long long fenr[MAXN];
void addl(long long n, long long x) {
  for (long long i = n; i < MAXN; i += (i & (-i))) fenl[i] += x;
}
void addr(long long n, long long x) {
  for (long long i = n; i < MAXN; i += (i & (-i))) fenr[i] += x;
}
long long getl(long long n) {
  long long ans = 0;
  for (long long i = n; i > 0; i -= (i & (-i))) ans += fenl[i];
  return ans;
}
long long getr(long long n) {
  long long ans = 0;
  for (long long i = n; i > 0; i -= (i & (-i))) ans += fenr[i];
  return ans;
}
int main() {
  std::ios::sync_with_stdio(0);
  ;
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long a[n];
  pair<long long, long long> p[n];
  long long b[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    p[i] = make_pair(a[i], i);
  }
  sort(p, p + n);
  long long ind = 1;
  for (long long i = 0; i < n; i++) {
    while (p[i].first == p[i + 1].first && i < n - 1) {
      b[p[i].second] = ind;
      i++;
    }
    b[p[i].second] = ind;
    ind++;
  }
  long long temp = 0;
  long long ans = 0;
  for (long long i = n - 1; i >= 0; i--) {
    temp += getr(b[i] - 1);
    addr(b[i], 1);
  }
  long long l = 0, r = 1;
  addr(b[0], -1);
  addl(b[0], 1);
  while (l != n) {
    while (temp > k && r < n) {
      addr(b[r], -1);
      temp -= getr(b[r] - 1);
      long long x = getl(b[r]);
      temp -= (l - x + 1);
      r++;
    }
    if (temp > k) break;
    ans += (n - r);
    if (l == n - 1) break;
    if (r == l + 1) {
      addr(b[r], -1);
      addl(b[r], 1);
      r++;
      l++;
    } else {
      l++;
      temp += getr(b[l] - 1);
      long long x = getl(b[l]);
      temp += (l - x);
      addl(b[l], 1);
    }
  }
  cout << ans;
  return 0;
}
