#include <bits/stdc++.h>
using namespace std;
long long int mod = 1e9 + 7;
int I_INF = 2e9;
long long int L_INF = 1e18;
void solve() {
  long long int n, i, j, l, r, index, ctr, num;
  cin >> n >> l >> r;
  vector<long long int> vect(n, 0);
  vect[0] = 2 * (n - 1);
  for (i = 1; i < n; i++) {
    vect[i] = 2 * (n - 1 - i);
    vect[i] += vect[i - 1];
  }
  vect[n - 1]++;
  index =
      (long long int)(lower_bound(vect.begin(), vect.end(), l) - vect.begin());
  ctr = (index > 0) ? vect[index - 1] : 0;
  while (ctr < r) {
    if (index == (n - 1)) {
      ctr++;
      if (ctr >= l && ctr <= r) cout << 1 << " ";
    } else {
      for (i = index + 1; i < n; i++) {
        ctr++;
        num = index + 1;
        if (ctr >= l && ctr <= r) cout << num << " ";
        ctr++;
        num = i + 1;
        if (ctr >= l && ctr <= r) cout << num << " ";
      }
    }
    index++;
  }
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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 int mod = 1e9 + 7;
int I_INF = 2e9;
long long int L_INF = 1e18;
void solve() {
  long long int n, i, j, l, r, index, ctr, num;
  cin >> n >> l >> r;
  vector<long long int> vect(n, 0);
  vect[0] = 2 * (n - 1);
  for (i = 1; i < n; i++) {
    vect[i] = 2 * (n - 1 - i);
    vect[i] += vect[i - 1];
  }
  vect[n - 1]++;
  index =
      (long long int)(lower_bound(vect.begin(), vect.end(), l) - vect.begin());
  ctr = (index > 0) ? vect[index - 1] : 0;
  while (ctr < r) {
    if (index == (n - 1)) {
      ctr++;
      if (ctr >= l && ctr <= r) cout << 1 << " ";
    } else {
      for (i = index + 1; i < n; i++) {
        ctr++;
        num = index + 1;
        if (ctr >= l && ctr <= r) cout << num << " ";
        ctr++;
        num = i + 1;
        if (ctr >= l && ctr <= r) cout << num << " ";
      }
    }
    index++;
  }
  cout << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```