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;
class Seg {
 public:
  int *a;
  Seg() { a = new int[131072 * 2 + 1]; }
  ~Seg() { delete[] a; }
  void insert(int x, int c) {
    for (int v = 131072 + x; v; v >>= 1) a[v] += c;
  }
  int query(int r) {
    if (r < 0) return 0;
    int s = 0;
    for (int v = 131072 + r; v; v >>= 1) {
      if (!(v & 1)) {
        s += a[v];
        if (v == (v & -v)) break;
        v--;
      }
    }
    return s;
  }
  int query(int l, int r) { return query(r) - query(l - 1); }
};
int n;
long long k;
int s[131072];
int scn, sc[131072];
long long solve() {
  long long cur = 0, sol = 0;
  Seg lseg, rseg;
  for (int i = n - 1; i > 0; i--) {
    rseg.insert(s[i], 1);
    cur += rseg.query(0, s[i] - 1);
  }
  for (int l = 0, r = 1; l < n; l++) {
    cur += lseg.query(s[l] + 1, n) + rseg.query(0, s[l] - 1);
    lseg.insert(s[l], 1);
    while (r <= l || r < n && cur > k) {
      cur -= lseg.query(s[r] + 1, n) + rseg.query(0, s[r] - 1);
      rseg.insert(s[r], -1);
      r++;
    }
    sol += n - r;
  }
  return sol;
}
int main(void) {
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    scanf("%d", s + i);
    sc[i] = s[i];
  }
  sort(sc, sc + n);
  scn = unique(sc, sc + n) - sc;
  for (int i = 0; i < n; i++) s[i] = lower_bound(sc, sc + scn, s[i]) - sc;
  cout << solve() << endl;
  return 0;
}
