Problem: You are given a complete directed graph K_n with n vertices: each pair of vertices u ≠ v in K_n have both directed edges (u, v) and (v, u); there are no self-loops.

You should find such a cycle in K_n that visits every directed edge exactly once (allowing for revisiting vertices).

We can write such cycle as a list of n(n - 1) + 1 vertices v_1, v_2, v_3, ..., v_{n(n - 1) - 1}, v_{n(n - 1)}, v_{n(n - 1) + 1} = v_1 — a visiting order, where each (v_i, v_{i + 1}) occurs exactly once.

Find the lexicographically smallest such cycle. It's not hard to prove that the cycle always exists.

Since the answer can be too large print its [l, r] segment, in other words, v_l, v_{l + 1}, ..., v_r.

Input

The first line contains the single integer T (1 ≤ T ≤ 100) — the number of test cases.

Next T lines contain test cases — one per line. The first and only line of each test case contains three integers n, l and r (2 ≤ n ≤ 10^5, 1 ≤ l ≤ r ≤ n(n - 1) + 1, r - l + 1 ≤ 10^5) — the number of vertices in K_n, and segment of the cycle to print.

It's guaranteed that the total sum of n doesn't exceed 10^5 and the total sum of r - l + 1 doesn't exceed 10^5.

Output

For each test case print the segment v_l, v_{l + 1}, ..., v_r of the lexicographically smallest cycle that visits every edge exactly once.

Example

Input


3
2 1 3
3 3 6
99995 9998900031 9998900031


Output


1 2 1 
1 3 2 3 
1 

Note

In the second test case, the lexicographically minimum cycle looks like: 1, 2, 1, 3, 2, 3, 1.

In the third test case, it's quite obvious that the cycle should start and end in vertex 1.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1.0);
const int M = 1e5 + 7;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long n, l, r, f = 0;
    cin >> n >> l >> r;
    l--;
    if (r == n * (n - 1) + 1) r--, f = 1;
    long long nl = (n - 1) * 2, nr = (n - 1) * 2;
    while (l >= nl && nl) l -= nl, nl -= 2;
    while (r >= nr && nr) r -= nr, nr -= 2;
    while (nl >= nr && nl) {
      long long tp = n - (nl / 2);
      long long nm = 0;
      vector<long long> v;
      for (long long i = tp + 1; i <= n; i++) v.push_back(tp), v.push_back(i);
      if (nl == nr)
        for (int i = l; i < r; i++) cout << v[i] << " ";
      else {
        for (int i = l; i < v.size(); i++) cout << v[i] << " ";
        l = 0;
      }
      nl -= 2;
    }
    if (f) cout << 1;
    cout << endl;
  }
  return 0;
}