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;
long long T, t, n, N, l, L, R;
vector<int> v;
pair<int, int> calcL() {
  for (n = l = 0; n < N; n++) {
    if ((l + (N - (n + 1))) > (L / 2)) break;
    l += (N - (n + 1));
  }
  return pair<int, int>(n, 1 + n + (int)((L / 2) - l));
}
void fill() {
  v.clear();
  pair<int, int> curr = calcL();
  for (n = L; n <= R; n++) {
    if (curr.first >= (N - 1)) {
      curr.first = 0;
    }
    if (curr.second >= N) {
      curr.second = 0;
    }
    if (n & 1) {
      v.push_back(curr.second);
      if (curr.second == (N - 1)) {
        curr = pair<int, int>((curr.first + 1), (curr.first + 2));
      } else
        curr.second++;
    } else
      v.push_back(curr.first);
  }
}
int main() {
  cin >> T;
  for (t = 0; t < T; t++) {
    cin >> N >> L >> R;
    L--;
    R--;
    fill();
    for (n = 0; n <= (R - L); n++) {
      printf("%d ", 1 + v[n]);
    }
    printf("\n");
  }
  return 0;
}