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;
int n;
long long l, r;
void out(long long sum, int now, int num, long long i) {
  if (now == n) {
    if (i != l) printf(" ");
    printf("1");
    return;
  }
  for (i; i <= min(sum + num - 1, r); i++) {
    int N;
    if ((i - sum + 1) % 2)
      N = now;
    else
      N = now + (i - sum + 1) / 2;
    if (i == l) {
      printf("%d", N);
    } else
      printf(" %d", N);
  }
}
int main() {
  scanf("%d", &T);
  while (T--) {
    long long sum = 1;
    scanf("%d%lld%lld", &n, &l, &r);
    long long now = 1, num = (n - now) * 2;
    for (now = 1; now <= n; now++) {
      num = num = (n - now) * 2;
      num = max(num, 1 * 1ll);
      if (sum <= l && l <= sum + num) {
        out(sum, now, num, l);
      } else if (l <= sum && sum + num <= r) {
        out(sum, now, num, sum);
      } else if (sum <= r && r <= sum + num) {
        out(sum, now, num, sum);
        break;
      }
      sum += num;
    }
    printf("\n");
  }
}