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 Min(long long a, long long b) { return (a < b) ? a : b; }
long long Max(long long a, long long b) { return (a < b) ? b : a; }
long long gcd(long long m, long long n) {
  if (n == 0) return m;
  return gcd(n, m % n);
}
long long lcm(long long m, long long n) { return m * n / gcd(m, n); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long tc;
  cin >> tc;
  while (tc--) {
    long long n, l, r;
    cin >> n >> l >> r;
    long long ff = l, cnt = r - l + 1, ai = -1;
    if (l > n * (n - 1)) {
      cout << "1\n";
      continue;
    }
    for (int i = 1; i < n; i++) {
      if (ff - 2 * (n - i) <= 0) {
        ai = i;
        break;
      }
      ff -= 2 * (n - i);
    }
    if (ai == -1) {
      ai = n - 1;
    }
    long long aj = ai + ff / 2 + (l & 1);
    l %= 2;
    while (cnt--) {
      if (l & 1) {
        cout << ai << ' ';
      } else {
        cout << aj << ' ';
        aj++;
        if (aj > n) {
          ai++;
          aj = ai + 1;
        }
        if (ai == n) {
          ai = 1;
        }
      }
      l = 1 - l;
    }
    cout << '\n';
  }
  exit(0);
}