#include <bits/stdc++.h>
using namespace std;
void brute_force(long long n) {
  set<int> graph[n + 1];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      graph[i].insert(j);
    }
  }
  int node = 1;
  vector<int> ans = {node};
  long long edges = n * (n - 1);
  while (edges != 0) {
    bool has_bridge = false;
    int bridge_neighbor = -1;
    int goto_neighbor = -1;
    for (auto neighbor : graph[node]) {
      if (graph[neighbor].size() < 1) {
        has_bridge = true;
        bridge_neighbor = neighbor;
      } else {
        goto_neighbor = neighbor;
        break;
      }
    }
    if (goto_neighbor == -1) {
      goto_neighbor = bridge_neighbor;
    }
    ans.push_back(goto_neighbor);
    graph[node].erase(goto_neighbor);
    node = goto_neighbor;
    edges--;
  }
  for (auto x : ans) {
    cout << x << " ";
  }
  cout << endl;
}
vector<int> ans;
void go(long long level, long long l, long long r, long long offset) {
  if (level == 0) return;
  long long prefix = (level - 1) * 2;
  if (l <= prefix) {
    long long idx = 1;
    for (int i = 2; i <= level; i++) {
      if (l <= idx && idx <= r) ans.push_back(1 + offset);
      idx++;
      if (l <= idx && idx <= r) ans.push_back(i + offset);
      idx++;
    }
  }
  if (r - prefix >= 1) {
    go(level - 1, l - prefix, r - prefix, offset + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long n, l, r;
    cin >> n >> l >> r;
    ans.clear();
    go(n, l, r, 0);
    if (r == n * (n - 1) + 1) ans.push_back(1);
    for (int i = 0; i < ans.size(); i++) {
      if (i) cout << " ";
      cout << ans[i];
    }
    cout << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following 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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void brute_force(long long n) {
  set<int> graph[n + 1];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      graph[i].insert(j);
    }
  }
  int node = 1;
  vector<int> ans = {node};
  long long edges = n * (n - 1);
  while (edges != 0) {
    bool has_bridge = false;
    int bridge_neighbor = -1;
    int goto_neighbor = -1;
    for (auto neighbor : graph[node]) {
      if (graph[neighbor].size() < 1) {
        has_bridge = true;
        bridge_neighbor = neighbor;
      } else {
        goto_neighbor = neighbor;
        break;
      }
    }
    if (goto_neighbor == -1) {
      goto_neighbor = bridge_neighbor;
    }
    ans.push_back(goto_neighbor);
    graph[node].erase(goto_neighbor);
    node = goto_neighbor;
    edges--;
  }
  for (auto x : ans) {
    cout << x << " ";
  }
  cout << endl;
}
vector<int> ans;
void go(long long level, long long l, long long r, long long offset) {
  if (level == 0) return;
  long long prefix = (level - 1) * 2;
  if (l <= prefix) {
    long long idx = 1;
    for (int i = 2; i <= level; i++) {
      if (l <= idx && idx <= r) ans.push_back(1 + offset);
      idx++;
      if (l <= idx && idx <= r) ans.push_back(i + offset);
      idx++;
    }
  }
  if (r - prefix >= 1) {
    go(level - 1, l - prefix, r - prefix, offset + 1);
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    long long n, l, r;
    cin >> n >> l >> r;
    ans.clear();
    go(n, l, r, 0);
    if (r == n * (n - 1) + 1) ans.push_back(1);
    for (int i = 0; i < ans.size(); i++) {
      if (i) cout << " ";
      cout << ans[i];
    }
    cout << endl;
  }
  return 0;
}

```