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 f(long long N, long long x) { return (2 * N - 1 - x) * x; }
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    long long N;
    cin >> N;
    long long l, r;
    cin >> l >> r;
    long long ng = 0, ok = N;
    while (ok - ng > 1) {
      long long mid = (ok + ng) / 2;
      if (f(N, mid) >= l)
        ok = mid;
      else
        ng = mid;
    }
    long long p = ok;
    long long stage = N - p;
    p--;
    long long cnt = (l - f(N, p) - 1) / 2,
              s = ((l - f(N, p) - 1) % 2 == 0 ? N - stage
                                              : N - stage + cnt + 1);
    for (int i = 0; i <= (r - l); i++) {
      if (i == r - l && r == N * (N - 1) + 1)
        cout << 1;
      else
        cout << s << " ";
      if (s == N - stage) {
        s = N - stage + cnt + 1;
      } else {
        if (cnt == stage - 1) {
          stage--;
          cnt = 0;
        } else
          cnt++;
        s = N - stage;
      }
    }
    cout << "\n";
  }
  return 0;
}