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;
void display(int lvl, int lt, int rt) {
  for (int i = lt; i <= rt; i++)
    printf("%d ", ((i & 1) ? lvl : lvl + (i >> 1)));
}
int main() {
  int t, n, i, j, k;
  cin >> t;
  while (t--) {
    long long l, r, lvl, sz, sc, mx;
    scanf("%d %lld %lld", &n, &l, &r);
    mx = n * (n - 1ll) + 1;
    if (l == mx) {
      printf("1\n");
      continue;
    }
    for (lvl = 1, sz = 0; lvl < n; lvl++) {
      sc = 2 * (n - lvl);
      if (sz >= l) {
        display(lvl, 1, min(sc, r - sz));
      } else if (sz + sc >= l) {
        display(lvl, l - sz, min(sc, r - sz));
      }
      sz += sc;
      if (sz >= r) {
        puts("");
        break;
      }
    }
    if (r == mx) {
      printf("1\n");
    }
  }
}