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 ar[200002];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  long long t, n, l, r;
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    if (l == n * (n - 1) + 1) {
      cout << "1\n";
      continue;
    }
    long long s = 0, j = 1;
    while (s + 2 * (n - j) < l) {
      s += 2 * (n - j);
      j++;
    }
    long long skipped = (l - s - 1) / 2;
    long long start = j, x = j + skipped + 1;
    ;
    long long end = n * (n - 1) + 1;
    for (long long a = l; a <= min(r, end - 1); a++) {
      if (a % 2) {
        cout << start << " ";
      } else {
        cout << x++ << " ";
      }
      if (x > n) {
        start++;
        x = start + 1;
      }
    }
    if (r == end) {
      cout << 1;
    }
    cout << "\n";
  }
  return 0;
}