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 main() {
  long long test;
  scanf("%lld", &test);
  while (test--) {
    long long i, j, k, l, r, n, m;
    scanf("%lld", &n);
    scanf("%lld", &l);
    scanf("%lld", &r);
    long long pre = 1, cub = 0, num;
    for (i = 1; i < n; i++) {
      if (l <= cub + 2 * (n - i)) {
        pre = i;
        break;
      } else
        cub += 2 * (n - i);
    }
    num = pre + 1;
    vector<long long> ans;
    for (i = cub + 1; i <= r && i <= n * (n - 1); i++) {
      if (i & 1)
        k = pre;
      else
        k = num++;
      if (i >= l) ans.push_back(k);
      if (num > n) {
        pre++;
        num = pre + 1;
      }
    }
    if (r == n * (n - 1) + 1) ans.push_back(1LL);
    for (long long bn : ans) {
      printf("%lld", bn);
      printf(" ");
    }
    printf("\n");
  }
}