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 MAX = 2e6 + 5, MOD = 1e9 + 7, MAXLG = log2(MAX) + 1;
const long long inf = 1e18 + 5;
int arr[MAX];
vector<long long> v;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    int len = r - l + 1;
    long long L = 0, R = 0;
    bool lt = false, rt = false;
    int lid = 0, rid = 0;
    for (int i = 1; i < n; i++) {
      long long val = 2ll * (n - i);
      if (!lt && L + val < l)
        L += val, lid = i;
      else
        lt = true;
      if (!rt && R + val < r)
        R += val, rid = i;
      else
        rt = true;
    }
    l -= (L + 1);
    lid++, rid++;
    deque<int> ans;
    for (int i = lid; i <= rid; i++) {
      for (int j = i + 1; j <= n; j++) {
        ans.push_back(i);
        ans.push_back(j);
      }
    }
    while (l--) ans.pop_front();
    while ((int)ans.size() > len) ans.pop_back();
    if ((int)ans.size() < len) ans.push_back(1);
    for (int x : ans) cout << x << " ";
    cout << "\n";
  }
}