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 = 1000 * 100 + 100;
struct fenwick {
  long long tree[maxn];
  fenwick() { memset(tree, 0, sizeof tree); }
  long long C(long long i) {
    long long ret = 0;
    for (; i > 0; i -= i & (-i)) ret += tree[i];
    return ret;
  }
  long long sum(int i, int j) { return C(j) - C(i - 1); }
  void update(long long x, long long val) {
    for (; x < maxn; x += x & (-x)) tree[x] += val;
  }
};
long long n, k, ans, l, r, x, y, z, t = 1, a[maxn], b[maxn], d[maxn], m;
map<long long, long long> mp;
fenwick f, c;
bool flag;
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> d[i];
    a[i] = d[i];
  }
  sort(d + 1, d + n + 1);
  for (int i = 1; i <= n; i++)
    if (!mp[d[i]]) mp[d[i]] = t++;
  for (int i = 1; i <= n; i++) b[i] = mp[a[i]];
  for (int i = 1; i <= n; i++) {
    x = f.sum(b[i] + 1, n);
    m += x;
    f.update(b[i], 1);
  }
  if (m <= k) {
    cout << (n * (n - 1)) / 2 << endl;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    x = 0;
    if (i != 1) {
      if (b[i - 1] != 1) m += f.sum(1, b[i - 1] - 1);
      m += c.sum(b[i - 1] + 1, n);
      c.update(b[i - 1], 1);
    }
    while (m > k) {
      x = 0;
      if (r == n) {
        flag = 1;
        break;
      }
      if (b[r + 1] != 1) m -= f.sum(1, b[r + 1] - 1);
      m -= c.sum(b[r + 1] + 1, n);
      f.update(b[++r], -1);
    }
    if (flag) break;
    if (i != 1) ans += (n - r);
  }
  cout << ans << endl;
  return 0;
}
