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;
struct BIT {
  vector<int> fen;
  int size;
  BIT() { size = 0; }
  BIT(int n) {
    fen.resize(n + 2);
    size = n;
  }
  void upd(int x, int val) {
    for (; x <= size; x += x & (-x)) fen[x] += val;
  }
  int get(int x) {
    int res = 0;
    for (; x > 0; x -= x & (-x)) res += fen[x];
    return res;
  }
} lft, rig;
const int MAX_N = 100000 + 20;
int b[MAX_N];
map<int, int> D;
int a[MAX_N];
long long ans;
long long inv, k;
int n;
void readInput() {
  cin >> n >> k;
  for (int i(0); i < int(n); ++i) {
    cin >> b[i];
    a[i] = b[i];
  }
}
inline int f(const int &x) { return (n + 1) - x; }
void erase(int id) {
  inv -= rig.get(a[id] - 1) + lft.get(f(a[id]) - 1);
  rig.upd(a[id], -1);
}
void add(int id) {
  inv += rig.get(a[id] - 1) + lft.get(f(a[id]) - 1);
  lft.upd(f(a[id]), 1);
}
int main() {
  ios::sync_with_stdio(false);
  readInput();
  {
    sort(b, b + n);
    int temp = 1;
    for (int i(0); i < int(n); ++i)
      if (D[b[i]] == 0) D[b[i]] = temp++;
    for (int i(0); i < int(n); ++i) a[i] = D[a[i]];
  }
  lft = rig = BIT(n);
  for (int i(n - 1); i >= int(0); --i) {
    inv += rig.get(a[i] - 1);
    rig.upd(a[i], 1);
  }
  int tail = 0;
  for (int i(0); i < int(n); ++i) {
    while (tail < n && tail <= i) erase(tail++);
    add(i);
    while (tail < n && inv > k) erase(tail++);
    ans += n - tail;
  }
  cout << ans << endl;
  return 0;
}
