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;
template <typename T>
inline T abs(T t) {
  return t < 0 ? -t : t;
}
const long long modn = 1000000007;
inline long long mod(long long x) { return x % modn; }
const int MAXN = 212345;
int n, m, k;
long long l, r;
int s[MAXN];
vector<int> ans;
long long stp;
void add(int x) {
  if (stp >= l && stp <= r) ans.push_back(x);
  stp++;
}
int main() {
  int t;
  scanf("%d", &t);
  for (int tt = 1; tt <= t; tt++) {
    ans.clear();
    scanf("%d%lld%lld", &n, &l, &r);
    stp = 1;
    int look = 2;
    while (look < n) {
      add(1);
      add(look);
      look++;
    }
    add(1);
    add(n);
    int base = 2;
    while (base < n) {
      add(base);
      if (stp > r) break;
      if (stp + 2ll * (n - 1 - base - 1 + 1) >= l) {
        for (int o = base + 1; o < n; o++) {
          add(o);
          add(base);
        }
      } else {
        stp += (long long)2ll * (n - 1 - base - 1 + 1);
      }
      add(n);
      base++;
    }
    add(1);
    for (int x : ans) printf("%d ", x);
    puts("");
  }
}