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;
long long l, r, curr = 1, n;
vector<long long> ans;
void rec(long long x) {
  if (x == n) {
    if (l <= curr && r >= curr) ans.push_back(1);
    return;
  }
  if (curr > r) return;
  if (curr + 2 * (n - x) - 1 < l) {
    curr += 2 * (n - x);
    rec(x + 1);
    return;
  }
  long long ptr = x + 1;
  for (int i = 0; i < 2 * (n - x); i++) {
    if (l <= curr && r >= curr) {
      if (i % 2 == 0)
        ans.push_back(x);
      else
        ans.push_back(ptr);
    }
    if (i % 2 == 1) ++ptr;
    ++curr;
  }
  rec(x + 1);
}
void solve() {
  cin >> n >> l >> r;
  curr = 1;
  ans.clear();
  rec(1);
  for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin.tie(NULL);
  int zz;
  cin >> zz;
  for (int zzz = 1; zzz <= zz; zzz++) {
    solve();
  }
  return 0;
}