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 gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int tcase;
  cin >> tcase;
  for (int tc = 1; tc <= tcase; tc++) {
    long long int n, l, r, count = 0, i = 0, j;
    cin >> n >> l >> r;
    while (count < l and count < n * (n - 1)) {
      count += 2 * (n - i - 1);
      i++;
    }
    if (count > l) {
      --i;
      count -= 2 * (n - i - 1);
    }
    if (count == n * (n - 1)) {
      if (l == count + 1)
        cout << "1\n";
      else if (l == r)
        cout << n << '\n';
      else
        cout << n << " 1\n";
    } else {
      long long int dif = l - count, start;
      count = 0;
      if (dif == 0) {
        cout << n << ' ';
        count++;
        start = ++i + 1;
      } else if (dif % 2)
        start = (++i) + dif / 2 + 1;
      else {
        start = (++i) + dif / 2;
        cout << start << ' ';
        start++;
        count++;
      }
      for (j = start; j <= n; j++) {
        if (count == r - l + 1) break;
        cout << i << ' ';
        count++;
        if (count == r - l + 1) break;
        cout << j << ' ';
        count++;
      }
      for (i = i + 1; i <= n; i++) {
        for (j = i + 1; j <= n; j++) {
          if (count == r - l + 1) break;
          cout << i << ' ';
          count++;
          if (count == r - l + 1) break;
          cout << j << ' ';
          count++;
        }
        if (count == r - l + 1) break;
      }
      if (count != r - l + 1) cout << "1";
      cout << '\n';
    }
  }
}