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 N = 100005;
const int inf = 1000 * 1000 * 1000;
const int mod = 998244353;
mt19937 myrand(chrono::steady_clock::now().time_since_epoch().count());
int t;
int n;
long long l, r;
int main() {
  cin >> t;
  while (t--) {
    scanf("%d%lld%lld", &n, &l, &r);
    vector<pair<long long, long long> > mas;
    vector<int> start;
    long long st = 1;
    for (int i = n - 1; i >= 1; i--) {
      mas.push_back(make_pair(st, st + 2 * i - 1));
      start.push_back(n - i);
      st += 2 * i;
    }
    mas.push_back(make_pair(st, st));
    for (int i = 0; i < (int)mas.size(); i++) {
      if (l > mas[i].second) continue;
      if (r < mas[i].first) continue;
      if (i == (int)mas.size() - 1) {
        printf("1");
        continue;
      }
      int ll = max(l, mas[i].first) - mas[i].first + 1;
      int rr = min(r, mas[i].second) - mas[i].first + 1;
      int s = start[i];
      for (int j = ll; j <= rr; j++) {
        if (j % 2) {
          printf("%d ", s);
        } else {
          printf("%d ", s + j / 2);
        }
      }
    }
    printf("\n");
  }
  return 0;
}