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;
const int MAXN = 110;
long long t, n, l, r, now;
vector<int> q;
int main() {
  cin >> t;
  while (t--) {
    while (q.size()) q.pop_back();
    cin >> n >> l >> r;
    now = 1;
    for (long long i = (1); i <= (n - 1); i++) {
      if (now > r) break;
      if (now + 2 * (n - i) - 1 < l) {
        now = now + 2 * (n - i);
        continue;
      }
      if (now >= l && now + 2 * (n - i) - 1 <= r) {
        for (long long j = (i + 1); j <= (n); j++)
          q.push_back(i), q.push_back(j);
      } else if (now >= l) {
        long long cnt = now;
        for (long long j = (i + 1); j <= (n); j++) {
          q.push_back(i);
          cnt++;
          if (cnt > r) break;
          q.push_back(j);
          cnt++;
          if (cnt > r) break;
        }
      } else if (now + 2 * (n - i) - 1 <= r) {
        long long cnt = now + 2 * (n - i) - 1;
        for (long long j = (n); j >= (i + 1); j--) {
          q.push_back(j);
          cnt--;
          if (cnt < l) break;
          q.push_back(i);
          cnt--;
          if (cnt < l) break;
        }
        reverse(q.begin(), q.end());
      } else {
        long long cnt = now;
        for (long long j = (i + 1); j <= (n); j++) {
          if (cnt >= l && cnt <= r) q.push_back(i);
          cnt++;
          if (cnt >= l && cnt <= r) q.push_back(j);
          cnt++;
        }
        break;
      }
      now = now + 2 * (n - i);
    }
    if (r == n * (n - 1) + 1) q.push_back(1);
    for (vector<int>::iterator it = q.begin(); it != q.end(); it++) {
      printf("%d ", *it);
    }
    printf("\n");
  }
  return 0;
}