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;
long long power(long long x, long long y, long long md = 1000000007LL) {
  long long res = 1;
  x %= md;
  while (y > 0) {
    if (y & 1) res = (res * x) % md;
    x *= x;
    if (x >= md) x %= md;
    y >>= 1;
  }
  return res;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    vector<long long> ans;
    long long prev = 1;
    for (long long i = 1; i <= n; i++) {
      long long st = prev;
      long long end = prev + 2 * (n - i) - 1;
      if (l > end) {
        prev = end + 1;
        continue;
      }
      if (r < st) break;
      long long st1 = max(l, st);
      long long en1 = min(r, end);
      for (long long j = st1; j <= en1; j++) {
        if ((j - st) % 2 == 0)
          ans.emplace_back(i);
        else {
          long long ex = (j - st + 1) / 2;
          ans.emplace_back(i + ex);
        }
      }
      prev = end + 1;
    }
    if (r == prev) ans.emplace_back(1);
    for (auto &x : ans) cout << x << " ";
    cout << '\n';
  }
}