#include <bits/stdc++.h>
using namespace std;
long long next() {
  long long x;
  cin >> x;
  return x;
}
const int maxn = 2e5 + 2, intf = 1e9;
const long long inf = 2e18, mod = 1e9;
const long double eps = 1e-6;
long long n;
void _print(long long bucket_id, long long l, long long r) {
  if (l > r) return;
  if (bucket_id == n) {
    cout << 1;
    return;
  }
  for (long long i = l; i <= r; i++) {
    if (i % 2 == 0) {
      cout << bucket_id << ' ';
    } else {
      cout << (i / 2) + 1 + bucket_id << ' ';
    }
  }
}
void solve() {
  long long l, r;
  cin >> n >> l >> r;
  l--, r--;
  long long sum = 0;
  for (long long bucket_id = 1; bucket_id <= n; bucket_id++) {
    long long cur_l = sum, cur_r = sum + 2 * (n - bucket_id) - 1;
    if (bucket_id == n) {
      if (r >= cur_l) {
        _print(bucket_id, 0, 1);
      }
      continue;
    }
    cur_l = max(cur_l, l) - sum;
    cur_r = min(cur_r, r) - sum;
    _print(bucket_id, cur_l, cur_r);
    sum += 2 * (n - bucket_id);
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  for (int t = next(); 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 next() {
  long long x;
  cin >> x;
  return x;
}
const int maxn = 2e5 + 2, intf = 1e9;
const long long inf = 2e18, mod = 1e9;
const long double eps = 1e-6;
long long n;
void _print(long long bucket_id, long long l, long long r) {
  if (l > r) return;
  if (bucket_id == n) {
    cout << 1;
    return;
  }
  for (long long i = l; i <= r; i++) {
    if (i % 2 == 0) {
      cout << bucket_id << ' ';
    } else {
      cout << (i / 2) + 1 + bucket_id << ' ';
    }
  }
}
void solve() {
  long long l, r;
  cin >> n >> l >> r;
  l--, r--;
  long long sum = 0;
  for (long long bucket_id = 1; bucket_id <= n; bucket_id++) {
    long long cur_l = sum, cur_r = sum + 2 * (n - bucket_id) - 1;
    if (bucket_id == n) {
      if (r >= cur_l) {
        _print(bucket_id, 0, 1);
      }
      continue;
    }
    cur_l = max(cur_l, l) - sum;
    cur_r = min(cur_r, r) - sum;
    _print(bucket_id, cur_l, cur_r);
    sum += 2 * (n - bucket_id);
  }
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(20);
  for (int t = next(); t--;) {
    solve();
  }
  return 0;
}

```