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;
int main() {
  cin.tie(0), ios_base::sync_with_stdio(0);
  int T;
  long long int i, j, p, q;
  cin >> T;
  long long int cnt2, N, L, R, cnt;
  for (int t = 0; t < T; t++) {
    cin >> N >> L >> R;
    cnt = 0;
    for (i = 1; i < N; i++) {
      if (L <= cnt + 2 * (N - i)) {
        for (j = L; j <= R - 1; j++) {
          if ((j - cnt) % 2 == 0)
            cnt2 = (j - cnt) / 2 + i;
          else
            cnt2 = i;
          cout << cnt2 << " ";
          if (cnt + 2 * (N - i) == j) cnt += 2 * (N - i), i++;
        }
        if (R <= N * (N - 1)) {
          if ((j - cnt) % 2 == 0)
            cnt2 = (j - cnt) / 2 + i;
          else
            cnt2 = i;
          cout << cnt2 << " ";
        }
        break;
      }
      cnt += 2 * (N - i);
      if (R < L) break;
    }
    if (R == N * (N - 1) + 1) {
      cout << 1;
    }
    cout << "\n";
  }
}