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 n, t, l, r, o, e, pos;
long long Find(long long k) {
  long long t = k;
  if (t % 2 == 1)
    t = (t + 1) / 2;
  else
    t /= 2;
  for (long long i = 0; i <= n - 1; i++) {
    if (n * i - i * (i + 1) / 2 < t && t <= n * (i + 1) - (i + 1) * (i + 2) / 2)
      return i + 1;
  }
}
int main() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld%lld%lld", &n, &l, &r);
    if (l % 2 == 0) {
      o = l + 1;
      e = l + 2;
      pos = Find(l);
    } else {
      o = l;
      e = l + 1;
    }
    long long s1 = Find(o);
    long long cnt = Find(e);
    long long s2 = e / 2 - ((cnt - 1) * n - (cnt - 1) * cnt / 2) + cnt;
    if (l % 2 == 0) {
      printf("%lld ", l / 2 - ((pos - 1) * n - (pos - 1) * pos / 2) + pos);
    }
    for (long long i = o; i <= r; i++) {
      if (i == n * (n - 1) + 1) {
        printf("1");
        break;
      }
      if (i % 2) {
        printf("%lld ", s1);
        if (s2 == n) s1++;
      } else {
        printf("%lld ", s2);
        s2++;
        if (s2 > n) s2 = s1 + 1;
      }
    }
    puts("");
  }
  return 0;
}