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 T;
long long N;
long long L, R;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> T;
  for (long long t = 0; t < T; t++) {
    cin >> N >> L >> R;
    if (L == N * (N - 1) + 1 && L == R) {
      cout << "1\n";
      continue;
    }
    long long ini = 0;
    long long s = 1;
    long long len = N - 1;
    while (ini + 2 * len < L) {
      ini += 2 * len;
      len--;
      s++;
    }
    ini++;
    long long first = s;
    long long second = s + 1;
    while (ini < L) {
      if (ini % 2 == 0) second++;
      if (second > N) first++, second = first + 1;
      ini++;
    }
    while (ini <= R && ini < N * (N - 1) + 1) {
      if (ini % 2)
        cout << first << " ";
      else
        cout << second << " ";
      if (ini % 2 == 0) second++;
      if (second > N) first++, second = first + 1;
      ini++;
    }
    if (R == N * (N - 1) + 1) cout << "1 ";
    cout << "\n";
  }
  return 0;
}