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 = 4 * atan(1);
const long long INF = 1e18;
const int MX = 100001;
int T, n;
long long l, r;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  while (T--) {
    cin >> n >> l >> r;
    long long first_elt = 1;
    long long num_visited = 0;
    long long length = 2 * (n - 1);
    while (num_visited + length < l) {
      num_visited += length;
      length -= 2;
      first_elt++;
      if (first_elt == n) {
        length = 1;
      }
    }
    vector<int> ans;
    int remaining = r - num_visited;
    while (ans.size() < remaining) {
      if (first_elt == n) {
        ans.push_back(1);
        break;
      }
      for (int i = first_elt + 1; i <= n; i++) {
        ans.push_back(first_elt);
        ans.push_back(i);
      }
      first_elt++;
    }
    for (long long i = 0; i < (long long)ans.size(); i++) {
      long long perm_index = num_visited + i + 1;
      if (perm_index >= l && perm_index < r) {
        cout << ans[i] << " ";
      }
      if (perm_index == r) {
        cout << ans[i] << "\n";
      }
    }
  }
}