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 int power_mod(long long int a, long long int x) {
  if (x == 0) return 1;
  long long int y = power_mod(a, x / 2);
  long long int ans = (y * y) % 1000000007;
  if (x % 2) ans = (ans * a) % 1000000007;
  return ans;
}
long long int inv(long long int a) { return power_mod(a, 1000000007 - 2); }
long long int power(long long int a, long long int x) {
  if (x == 0) return 1;
  long long int y = power(a, x / 2);
  long long int ans = (y * y);
  if (x % 2) ans *= a;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long int n, l, r;
    cin >> n >> l >> r;
    long long int curr = 1;
    vector<long long int> v(n + 1);
    long long int sum = 0;
    long long int index = 1;
    vector<long long int> ans;
    for (long long int i = 1; i <= n - 1; i++) {
      if (sum + 2 * (n - i) >= l) {
        long long int curr = i;
        long long int curr_next = i + 1;
        long long int count = 2 * (n - i);
        for (long long int j = sum + 1; j <= min(n * (n - 1), r); j++) {
          if (j >= l) {
            ans.push_back(curr);
          }
          count--;
          if (count == 0) {
            i++;
            curr = i;
            curr_next = i + 1;
            count = 2 * (n - i);
            continue;
          }
          j++;
          if (j > r) break;
          if (j >= l) {
            ans.push_back(curr_next);
          }
          curr_next++;
          count--;
          if (count == 0) {
            i++;
            curr = i;
            curr_next = i + 1;
            count = 2 * (n - i);
            continue;
          }
        }
        break;
      } else {
        sum = sum + (2 * (n - i));
      }
    }
    if (r == (n * (n - 1)) + 1) ans.push_back(1);
    for (auto i : ans) cout << i << " ";
    cout << endl;
  }
}