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 = 3e5 + 5, K = 11, M = N * 4;
const long long MOD = 998244353, oo = 1e9, OO = 1e18, mod = MOD;
const double pi = acos(-1), eps = 1e-17;
long long di[] = {0, 0, 1, -1};
long long dj[] = {1, -1, 0, 0};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n, l, r;
    cin >> n >> l >> r;
    long long now = 0, st = 1;
    long long sum = l;
    for (long long i = n - 1; i > 0; i--) {
      now = i * 2;
      l -= now;
      r -= now;
      if (l <= 0) {
        l += now;
        r += now;
        sum = i;
        break;
      }
      st++;
    }
    long long ans = 0;
    long long p = 0;
    for (long long i = sum; "Hello"; i--) {
      long long tmp = i * 2;
      long long dor = st;
      if (st == n) dor = 1, tmp = 1;
      long long cnt = st + 1;
      while (tmp--) {
        ans++;
        if (ans >= l && ans <= r) {
          cout << dor << " ";
        }
        if (dor == st)
          dor = cnt++;
        else
          dor = st;
        if (ans > r) goto A;
      }
      st++;
      if (ans > r) goto A;
    }
  A:
    cout << endl;
  }
}