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 long long N = 200001;
const long long MOD = 1000000007;
const long long INF = LLONG_MAX;
vector<long long> v;
void pre() {
  v.push_back(1LL);
  for (long long i = 1; i <= 1e5; i++) v.push_back(i * 2);
  for (long long i = 1; i < v.size(); i++) v[i] += v[i - 1];
}
long long Getfirst(long long n, long long x) {
  long long id = upper_bound(v.begin(), v.end(), x) - v.begin();
  return (n - id);
}
long long Getlast(long long n, long long x) {
  long long id = upper_bound(v.begin(), v.end(), x) - v.begin() - 1;
  long long dif = x - v[id] + 1;
  return (n - dif / 2);
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  pre();
  int tc = 1, cs = 0;
  cin >> tc;
  while (tc--) {
    long long n, l, r, i, limit;
    cin >> n >> l >> r;
    limit = n * (n - 1) + 1;
    vector<long long> ans;
    for (i = l; i <= r; i++) {
      if (i == limit) {
        ans.push_back(1);
        break;
      }
      if (i & 1)
        ans.push_back(Getfirst(n, limit - i));
      else
        ans.push_back(Getlast(n, limit - i));
    }
    for (auto it : ans) cout << it << ' ';
    cout << '\n';
    ;
  }
}