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;
void solve() {
  int n;
  long long l, r, len;
  cin >> n >> l >> r;
  l--, r--;
  len = (r - l + 1);
  int i = 0;
  for (; i < n; i++) {
    if ((n - i - 1) * 2 <= l) {
      l -= (n - i - 1) * 2;
      r -= (n - i - 1) * 2;
    } else {
      break;
    }
  }
  vector<int> ans;
  for (int j = l / 2 + i + 1; j < n; j++) {
    ans.push_back(i);
    ans.push_back(j);
  }
  r -= l / 2 * 2;
  l -= l / 2 * 2;
  for (i++; i < n && ans.size() <= r; i++) {
    for (int j = i + 1; j < n && ans.size() <= r; j++) {
      ans.push_back(i);
      ans.push_back(j);
    }
  }
  ans.push_back(0);
  for (int i = l; i <= r; i++) {
    cout << ans[i] + 1 << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin.sync_with_stdio(0);
  cout.tie(0);
  cout.sync_with_stdio(0);
  cout.precision(9);
  int q;
  cin >> q;
  while (q--) {
    solve();
  }
  return 0;
}