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;
const int N = 300005;
int main() {
  long long test = 1;
  scanf("%d", &test);
  while (test--) {
    long long n;
    long long l, r;
    scanf("%lld %lld %lld", &n, &l, &r);
    if (l == 1LL * n * (n - 1) + 1) {
      printf("1\n");
      continue;
    }
    long long start = 1;
    long long curr = 0;
    while ((n - start) && curr + 2 * (n - start) < l) {
      curr += 2 * (n - start);
      start++;
    }
    long long idx = start + (l - curr + 1) / 2;
    long long tot = r - l + 1;
    vector<long long> v;
    if ((l - curr) % 2 == 0) {
      v.push_back(idx);
      idx++;
    }
    for (long long j = idx; j <= n && v.size() < tot; j++) {
      v.push_back(start);
      if (v.size() < tot) {
        v.push_back(j);
      }
    }
    for (long long i = start + 1; i <= n && v.size() < tot; i++) {
      for (long long j = i + 1; j <= n && v.size() < tot; j++) {
        v.push_back(i);
        if (v.size() < tot) {
          v.push_back(j);
        }
      }
    }
    if (v.size() < tot) {
      v.push_back(1);
    }
    for (long long i = 0; i < v.size(); i++) {
      printf("%lld ", v[i]);
    }
    printf("\n");
  }
  return 0;
}