#include <bits/stdc++.h>
using namespace std;
long long int max(long long int a, long long int b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a <= b)
    return a;
  else
    return b;
}
void solve() {
  long long int n, l, r;
  cin >> n >> l >> r;
  if (l == (1LL * n * (n - 1)) + 1) {
    cout << 1 << endl;
    return;
  }
  long long int sum = 0;
  long long int i;
  for (i = n - 1; i >= 1; i--) {
    sum += 1LL * 2 * i;
    if (sum >= l) {
      break;
    }
  }
  vector<long long int> ans;
  long long int total = r - l + 1;
  if (r == (1LL * n * (n - 1)) + 1) total--;
  long long int count = sum - l + 1;
  long long int a, b;
  if ((count % 2) == 0) {
    a = n - i;
    b = n - (count / 2) + 1;
  } else {
    ans.push_back(n - ((count) / 2));
    a = n - i;
    b = n - (count / 2) + 1;
    total--;
  }
  if (count == 1) {
    a++;
    b = a + 1;
  }
  while (total > 0) {
    if (b == n + 1) {
      a++;
      b = a + 1;
    }
    ans.push_back(a);
    total--;
    if (total == 0) break;
    ans.push_back(b);
    total--;
    b++;
  }
  if (r == (1LL * n * (n - 1)) + 1) ans.push_back(1);
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 max(long long int a, long long int b) {
  if (a >= b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a <= b)
    return a;
  else
    return b;
}
void solve() {
  long long int n, l, r;
  cin >> n >> l >> r;
  if (l == (1LL * n * (n - 1)) + 1) {
    cout << 1 << endl;
    return;
  }
  long long int sum = 0;
  long long int i;
  for (i = n - 1; i >= 1; i--) {
    sum += 1LL * 2 * i;
    if (sum >= l) {
      break;
    }
  }
  vector<long long int> ans;
  long long int total = r - l + 1;
  if (r == (1LL * n * (n - 1)) + 1) total--;
  long long int count = sum - l + 1;
  long long int a, b;
  if ((count % 2) == 0) {
    a = n - i;
    b = n - (count / 2) + 1;
  } else {
    ans.push_back(n - ((count) / 2));
    a = n - i;
    b = n - (count / 2) + 1;
    total--;
  }
  if (count == 1) {
    a++;
    b = a + 1;
  }
  while (total > 0) {
    if (b == n + 1) {
      a++;
      b = a + 1;
    }
    ans.push_back(a);
    total--;
    if (total == 0) break;
    ans.push_back(b);
    total--;
    b++;
  }
  if (r == (1LL * n * (n - 1)) + 1) ans.push_back(1);
  for (auto it : ans) {
    cout << it << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```