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 main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n, l, r;
    scanf("%lld%lld%lld", &n, &l, &r);
    long long suml = 0, sumr = 0;
    long long headl = 1, headr = -1;
    long long flag1 = 0;
    for (long long i = 1; i < n; ++i) {
      if (!flag1) {
        suml += 2 * (n - i);
      }
      sumr += 2 * (n - i);
      if (suml >= l && !flag1) {
        suml -= 2 * (n - i);
        flag1 = 1;
        headl = i;
      } else if (sumr >= r) {
        headr = i;
        break;
      }
    }
    long long vv = 0;
    if (headr == -1) {
      vv = 1;
    }
    long long flag = 1;
    long long now = headl + 1;
    for (long long i = suml + 1; i <= sumr; ++i) {
      if (flag == 1) {
        if (l <= i && i <= r) {
          printf("%lld ", headl);
        }
        flag = 2;
      } else if (flag == 2) {
        if (l <= i && i <= r) {
          printf("%lld ", now);
        }
        now++;
        if (now > n) {
          headl++;
          now = headl + 1;
        }
        flag = 1;
      }
    }
    if (vv == 1) {
      printf("1 ");
    }
    putchar('\n');
  }
  return 0;
}