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;
void upgrade() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
}
int main() {
  upgrade();
  int tc;
  cin >> tc;
  while (tc--) {
    long long n, l, r;
    cin >> n >> l >> r;
    long long add = 2 * n - 2, cnt = 0, h = 1;
    while (add != 0 && cnt + add < l) {
      cnt += add;
      add -= 2;
      h++;
    }
    if (add == 0) {
      cout << 1 << '\n';
      continue;
    }
    long long diff = l - cnt;
    bool f = (diff % 2) == 1;
    long long st = (diff - 1) / 2 + h + 1;
    for (int i = 0; i < r - l + 1; i++) {
      if (f) {
        cout << h << ' ';
      } else {
        cout << st << ' ';
        if (st == n) {
          h++;
          st = h + 1;
          if (h == n && i != r - l) {
            cout << 1;
            break;
          }
        } else
          st++;
      }
      f = !f;
    }
    cout << '\n';
  }
}