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;
const long long INF = 1e17 + 9;
long long n, l, r;
void calc(long long ly, long long faltl, long long faltr) {
  if (ly == n) {
    r = 1;
  } else if (faltl >= 2 * n - ly * 2) {
    calc(ly + 1, faltl - 2 * n + ly * 2, faltr);
  } else {
    long long ind = ly + 1;
    long long guard = 0;
    for (long long i = 0; i < 2 * n - ly * 2 && faltl > 0; i++) {
      if (i % 2) ind++;
      faltl--;
      guard = i + 1;
    }
    for (long long i = guard; i < 2 * n - ly * 2 && faltr > 0; i++) {
      if (i % 2)
        cout << ind++ << " ";
      else
        cout << ly << " ";
      faltr--;
    }
    if (faltr > 0 && ly < n - 1) {
      calc(ly + 1, faltl, faltr);
    } else {
      r = faltr;
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    calc(1, l - 1, r - l + 1);
    if (r == 1)
      cout << 1 << '\n';
    else
      cout << '\n';
  }
  return 0;
}