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, n, l, r;
void stampa(int x, int y) {
  if (r == 0) return;
  cout << x << " ";
  r--;
  if (x == n && y >= n) return;
  int succ;
  if (y > x) {
    succ = x + (y == n);
  } else {
    if (x == n) {
      succ = y + 1;
    } else {
      succ = x + 1;
    }
  }
  stampa(y, succ);
}
int main() {
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    r = (r - l + 1);
    for (long long i = 1; i <= n; i++) {
      if (l > ((n - i) * 2ll)) {
        l -= ((n - i) * 2ll);
      } else {
        if (l % 2 == 0)
          stampa(i + l / 2, i + ((i + l / 2 == n)));
        else
          stampa(i, i + (l + 1) / 2);
        break;
      }
    }
    if (r > 0) cout << "1";
    cout << "\n";
  }
  return 0;
}