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;
struct R {
  long long int nd, i;
};
int main() {
  long long int n, l, r, st;
  scanf("%*d");
  while (~scanf("%lld %lld %lld", &n, &l, &r)) {
    long long int p, i, j, nd;
    vector<R> v;
    v.push_back({1, 1});
    for (p = 1, i = 2, j = n - 1; i < n; i++, j--) v.push_back({i, p += j + j});
    long long int st = 0, en = v.size() - 1, md, pos = -1;
    while (st <= en) {
      md = (st + en) / 2;
      if (v[md].i > l)
        en = md - 1;
      else {
        pos = max(pos, md);
        st = md + 1;
      }
    }
    vector<R> pet;
    for (nd = v[pos].nd, i = v[pos].i; i <= r && nd < n; nd++) {
      for (j = nd + 1; j <= n && i <= r; j++)
        pet.push_back({nd, i++}), pet.push_back({j, i++});
    }
    if (nd == n) pet.push_back({1, i++});
    for (auto it : pet)
      if (it.i >= l && it.i <= r) printf("%lld ", it.nd);
    puts("");
  }
  return 0;
}