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;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    if (n == 1) {
      cout << "1" << endl;
      continue;
    }
    long long x = 2 * (n - 1);
    long long c = 0;
    r = r - l + 1;
    while (x > 0 && l - x > 0) {
      l -= x;
      c++;
      x -= 2;
    }
    long long f;
    for (long long i = c + 1;; i++) {
      f = 0;
      long long k = i;
      if (i == c + 1) {
        if (l % 2 == 1) {
          k += (l + 1) / 2;
        } else {
          k += l / 2;
          f = 1;
        }
      } else
        k++;
      while (k <= n) {
        if (!f)
          cout << i << " ";
        else {
          cout << k << " ";
          k++;
        }
        r--;
        if (r == 0) break;
        f = !f;
      }
      if (r == 0) break;
      if (i == n) {
        cout << "1";
        break;
      }
    }
    cout << endl;
  }
}