#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const int N = 100005;
void solve() {
  long long n, s, e;
  cin >> n >> s >> e;
  if (s == n * n - n + 1) {
    cout << 1 << '\n';
    return;
  } else {
    long long a = n * 2 - 2;
    long long sq = sqrt((a + 1) * (a + 1) - 4 * s);
    long long st = (a + 1 - sq) / 2;
    if (sq * sq < (a + 1) * (a + 1) - 4 * s) st++;
    long long prev = (st - 1) * (a - (st - 2));
    if (prev >= s) {
      st--;
      prev = (st - 1) * (a - (st - 2));
    }
    long long l = st, r = (s - prev + 1) / 2 + st;
    if ((s % 2) == 0) {
      cout << r << " ";
      r++;
      s++;
    }
    while (s <= min(e, n * n - n)) {
      if (r > n) {
        l++;
        r = l + 1;
      }
      cout << l << " ";
      s++;
      if (s <= e)
        cout << r << " ";
      else
        break;
      r++;
      s++;
    }
    if (e == n * n - n + 1) {
      cout << 1 << '\n';
    } else
      cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
}

### Prompt

Generate 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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
const int N = 100005;
void solve() {
  long long n, s, e;
  cin >> n >> s >> e;
  if (s == n * n - n + 1) {
    cout << 1 << '\n';
    return;
  } else {
    long long a = n * 2 - 2;
    long long sq = sqrt((a + 1) * (a + 1) - 4 * s);
    long long st = (a + 1 - sq) / 2;
    if (sq * sq < (a + 1) * (a + 1) - 4 * s) st++;
    long long prev = (st - 1) * (a - (st - 2));
    if (prev >= s) {
      st--;
      prev = (st - 1) * (a - (st - 2));
    }
    long long l = st, r = (s - prev + 1) / 2 + st;
    if ((s % 2) == 0) {
      cout << r << " ";
      r++;
      s++;
    }
    while (s <= min(e, n * n - n)) {
      if (r > n) {
        l++;
        r = l + 1;
      }
      cout << l << " ";
      s++;
      if (s <= e)
        cout << r << " ";
      else
        break;
      r++;
      s++;
    }
    if (e == n * n - n + 1) {
      cout << 1 << '\n';
    } else
      cout << '\n';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
}

```