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 n, l, r;
int Go() {
  scanf("%lld %lld %lld", &n, &l, &r);
  long long len = r - l + 1;
  long long x = n;
  long long y = 0;
  long long start = 0;
  while (1) {
    y += 2 * (x - 1);
    x--;
    start++;
    if (x <= 0 || y >= l) break;
  }
  long long k = y - 2 * x;
  long long d = (l - k + 1) / 2;
  long long add = start + d;
  vector<long long> Ans;
  long long count = 0;
  if (l % 2 == 0) {
    Ans.push_back(add++);
    count++;
  }
  while (count < len) {
    if (add == n + 1) {
      start++;
      add = start + 1;
    }
    Ans.push_back(start);
    count++;
    if (count == len) break;
    Ans.push_back(add);
    count++;
    add++;
  }
  if (r == n * (n - 1) + 1) {
    Ans.pop_back();
    Ans.push_back(1);
  }
  for (int i = 0; i < Ans.size(); ++i) printf("%lld ", Ans[i]);
  puts("");
  return 0;
}
int main() {
  int t;
  if (1)
    scanf("%d", &t);
  else
    t = 1;
  while (t--) Go();
}