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, n, s, lc;
long long l, r, k;
inline long long read() {
  long long ans = 0;
  char c = getchar();
  while (c < 48 || c > 57) c = getchar();
  while (c >= 48 && c <= 57)
    ans = (ans << 3) + (ans << 1) + (c ^ 48), c = getchar();
  return ans;
}
inline void write(int x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + 48);
}
int main() {
  t = read();
  while (t--) {
    n = read(), l = read(), r = read(), k = 0;
    for (register int i = 1; i < n; ++i) {
      s = (n - i) << 1;
      if (k + s < l) {
        k += s;
        continue;
      }
      if (k < l)
        lc = l - k - 1, k = l - 1;
      else
        lc = 0;
      for (register int j = lc; j < s - 1; ++j) {
        ++k;
        if (k > r) break;
        if (!(j & 1))
          write(i);
        else
          write((j >> 1) + i + 1);
        putchar(' ');
      }
      if (k > r) break;
      ++k;
      if (k > r) break;
      write(n), putchar(' ');
    }
    if (k < r) putchar(49);
    putchar('\n');
  }
  return 0;
}