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() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    long long len = n * (n - 1) + 1;
    long long i = l;
    for (; i <= r && i <= 2 * (n - 2) + 1; i++) {
      if (i % 2)
        cout << 1 << " ";
      else
        cout << 1 + i / 2 << " ";
    }
    if (i > r) {
      cout << "\n";
      continue;
    }
    map<long long, tuple<long long, long long, long long>> seen;
    long long si = 2 * (n - 2) + 2;
    for (long long k = 2; k <= n - 1; k++) {
      long long len = 2 * (n - k);
      seen[si] = make_tuple(k, len, si + len - 1);
      si += len;
    }
    long long idx = i;
    while (!(seen.count(idx) > 0) && seen.size() != 0) {
      idx--;
    }
    while (i <= r && i <= len - 2) {
      for (; i <= r && i <= get<2>(seen[idx]); i++) {
        long long j = i - idx;
        if (j == 0)
          cout << n << " ";
        else if (j % 2 == 0)
          cout << get<0>(seen[idx]) + (j / 2) << " ";
        else
          cout << get<0>(seen[idx]) << " ";
      }
      idx = i;
    }
    if (i <= r && r <= len) {
      if (i == len - 1 && i <= r) {
        cout << n << " ";
        i++;
      }
      if (i == len && i <= r) cout << 1 << " ";
    }
    cout << "\n";
  }
  return 0;
}