#include <bits/stdc++.h>
using namespace std;
long long n, t, l, r, start, star, have;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    have = 0;
    star = start = n + 1;
    for (long long i = 1; i <= n; i++)
      if (2LL * n * i - i * (i + 1LL) >= l) {
        start = i;
        have = 2LL * n * (i - 1LL) - (i - 1LL) * i;
        break;
      }
    for (long long i = start + 1; i <= n; i++)
      if (have + 2 < l)
        have += 2LL;
      else {
        have += 2LL;
        if (have == l) {
          cout << i << " ";
          l++;
          star = i + 1;
        } else
          star = i;
        break;
      }
    for (long long i = start; i <= n; i++)
      for (long long j = (i == start) ? star : i + 1; j <= n; j++) {
        if (l == r) {
          cout << i << " ";
          i = n + 1;
          break;
        }
        if (l > r) {
          i = n + 1;
          break;
        }
        l += 2;
        cout << i << " " << j << " ";
      }
    if (r == n * (n - 1LL) + 1) cout << "1";
    cout << "\n";
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, t, l, r, start, star, have;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n >> l >> r;
    have = 0;
    star = start = n + 1;
    for (long long i = 1; i <= n; i++)
      if (2LL * n * i - i * (i + 1LL) >= l) {
        start = i;
        have = 2LL * n * (i - 1LL) - (i - 1LL) * i;
        break;
      }
    for (long long i = start + 1; i <= n; i++)
      if (have + 2 < l)
        have += 2LL;
      else {
        have += 2LL;
        if (have == l) {
          cout << i << " ";
          l++;
          star = i + 1;
        } else
          star = i;
        break;
      }
    for (long long i = start; i <= n; i++)
      for (long long j = (i == start) ? star : i + 1; j <= n; j++) {
        if (l == r) {
          cout << i << " ";
          i = n + 1;
          break;
        }
        if (l > r) {
          i = n + 1;
          break;
        }
        l += 2;
        cout << i << " " << j << " ";
      }
    if (r == n * (n - 1LL) + 1) cout << "1";
    cout << "\n";
  }
}

```