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 q, n, l, r, sum[100005];
signed main() {
  cin >> q;
  while (q--) {
    cin >> n >> l >> r;
    long long s = n * (n - 1), p = 0;
    if (l == s + 1) {
      puts("1");
      continue;
    }
    sum[0] = 0;
    for (long long i = 1; i < n; i++) sum[i] = sum[i - 1] + 2 * (n - i);
    for (p = 1; p < n; p++)
      if (sum[p] >= l) break;
    if (r < s + 1)
      for (long long i = l; i <= r; i++) {
        if (i % 2 == 1)
          cout << p << ' ';
        else
          cout << p + (i - sum[p - 1]) / 2 << ' ';
        if (i == sum[p]) p++;
      }
    else {
      for (long long i = l; i < r; i++) {
        if (i % 2 == 1)
          cout << p << ' ';
        else
          cout << p + (i - sum[p - 1]) / 2 << ' ';
        if (i == sum[p]) p++;
      }
      cout << 1;
    }
    puts("");
  }
  return 0;
}