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 a[1000006];
long long n;
map<long long, long long> mp;
int main() {
  long long Q;
  cin >> Q;
  while (Q--) {
    cin >> n;
    long long l, r;
    cin >> l >> r;
    a[0] = (n - 1) * 2;
    long long k = a[0] - 2;
    long long po = 0;
    for (int i = 1; i < 1000006; i++) {
      a[i] = a[i - 1] + k;
      k -= 2;
      if (k == 0) {
        po = i + 1;
        break;
      }
    }
    long long qq, q = lower_bound(a, a + po, l) - a;
    if (q == 0)
      qq = 0;
    else
      qq = a[q - 1] + 1;
    long long qqq = a[q] + 1;
    q++;
    while (l <= r) {
      if (l == n * (n - 1) + 1) {
        cout << 1;
        break;
      }
      if (l == qqq) {
        q = lower_bound(a, a + po, l) - a;
        qq = a[q - 1] + 1;
        qqq = a[q] + 1;
        q++;
        cout << q << ' ';
        l++;
        continue;
      }
      if (l % 2 == 1)
        cout << q << ' ';
      else
        cout << (l - qq + 1) / 2 + q << ' ';
      l++;
    }
    cout << endl;
  }
  return 0;
}