#include <bits/stdc++.h>
using namespace std;
vector<int> func(int n, long long l, long long r) {
  long long start = 1;
  long long end = 2 * 1ll * (n - 1);
  vector<int> ans;
  if (start <= l && l <= end) {
    while (l <= r && l <= end) {
      if (l % 2 == 1) {
        ans.push_back(1);
      } else {
        ans.push_back(l / 2 + 1);
      }
      l++;
    }
  }
  for (int i = 2; i < n; ++i) {
    if (i == n - 1) {
      start = end + 1;
      end = start + 1;
      if (start <= l && l <= end) {
        while (l <= r && l <= end) {
          if (start == l) {
            ans.push_back(n - 1);
          } else {
            ans.push_back(n);
          }
          l++;
        }
      }
    } else {
      long long tmp_start = end;
      start = end + 1;
      end = start + 2 + (n - i - 1) * 1ll * 2;
      end = end - 1;
      if (start <= l && l <= end) {
        while (l <= r && l <= end) {
          int ref_value = l - tmp_start;
          if (ref_value == 1) {
            ans.push_back(i);
          } else if (l != end) {
            if (ref_value % 2 == 0) {
              ans.push_back(i + ref_value / 2);
            } else {
              ans.push_back(i);
            }
          } else {
            ans.push_back(n);
          }
          l++;
        }
      }
    }
  }
  start = end + 1;
  end = end + 1;
  if (start <= l && l <= end) {
    while (l <= r && l <= end) {
      ans.push_back(1);
      l++;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    long long l, r;
    cin >> n >> l >> r;
    vector<int> v = func(n, l, r);
    for (int x : v) {
      cout << x << " ";
    }
    cout << endl;
  }
  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;
vector<int> func(int n, long long l, long long r) {
  long long start = 1;
  long long end = 2 * 1ll * (n - 1);
  vector<int> ans;
  if (start <= l && l <= end) {
    while (l <= r && l <= end) {
      if (l % 2 == 1) {
        ans.push_back(1);
      } else {
        ans.push_back(l / 2 + 1);
      }
      l++;
    }
  }
  for (int i = 2; i < n; ++i) {
    if (i == n - 1) {
      start = end + 1;
      end = start + 1;
      if (start <= l && l <= end) {
        while (l <= r && l <= end) {
          if (start == l) {
            ans.push_back(n - 1);
          } else {
            ans.push_back(n);
          }
          l++;
        }
      }
    } else {
      long long tmp_start = end;
      start = end + 1;
      end = start + 2 + (n - i - 1) * 1ll * 2;
      end = end - 1;
      if (start <= l && l <= end) {
        while (l <= r && l <= end) {
          int ref_value = l - tmp_start;
          if (ref_value == 1) {
            ans.push_back(i);
          } else if (l != end) {
            if (ref_value % 2 == 0) {
              ans.push_back(i + ref_value / 2);
            } else {
              ans.push_back(i);
            }
          } else {
            ans.push_back(n);
          }
          l++;
        }
      }
    }
  }
  start = end + 1;
  end = end + 1;
  if (start <= l && l <= end) {
    while (l <= r && l <= end) {
      ans.push_back(1);
      l++;
    }
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    long long l, r;
    cin >> n >> l >> r;
    vector<int> v = func(n, l, r);
    for (int x : v) {
      cout << x << " ";
    }
    cout << endl;
  }
  return 0;
}

```