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;
void yes() { cout << "YES" << endl; }
void no() { cout << "NO" << endl; }
void solve(int qq) {
  long long n;
  cin >> n;
  long long l, r;
  cin >> l >> r;
  long long cnt = 0;
  long long odd = 0;
  long long start = 0;
  if (l == n * (n - 1) + 1) {
    cout << 1 << endl;
    return;
  }
  for (long long i = 1; i < n; i++) {
    cnt += (n - i) * 2;
    if (l <= cnt) {
      odd = i;
      start = cnt - (n - i) * 2 + 1;
      break;
    }
  }
  long long even = odd + ((l - start) / 2 + 1);
  for (long long i = l; i <= r; i++) {
    if (i % 2 == 0) {
      cout << even << " ";
      even++;
      if (even > n) {
        odd++;
        even = odd + 1;
        if (odd == n) {
          odd = 1;
        }
      }
    } else {
      cout << odd << " ";
    }
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve(i);
  }
  return 0;
}