#include <bits/stdc++.h>
const int N = 3 * 1e5 + 5;
const long long MOD = 1000000007;
const long long inf = 1e18;
using namespace std;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long ctr = 0;
  long long lt = l;
  if (l == n * (n - 1) + 1) {
    cout << 1 << "\n";
    return;
  }
  while (lt > 0) {
    ctr++;
    lt -= 2 * (n - ctr);
  }
  lt += 2 * (n - ctr);
  long long val = (r - l + 1);
  while (val) {
    val--;
    if (ctr == n) {
      cout << "1"
           << " "
           << "\n";
      return;
    }
    if (lt % 2 == 1)
      cout << ctr;
    else
      cout << ctr + lt / 2;
    cout << " ";
    if (lt < 2 * (n - ctr))
      lt++;
    else {
      ctr++;
      lt = 1;
    }
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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>
const int N = 3 * 1e5 + 5;
const long long MOD = 1000000007;
const long long inf = 1e18;
using namespace std;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long ctr = 0;
  long long lt = l;
  if (l == n * (n - 1) + 1) {
    cout << 1 << "\n";
    return;
  }
  while (lt > 0) {
    ctr++;
    lt -= 2 * (n - ctr);
  }
  lt += 2 * (n - ctr);
  long long val = (r - l + 1);
  while (val) {
    val--;
    if (ctr == n) {
      cout << "1"
           << " "
           << "\n";
      return;
    }
    if (lt % 2 == 1)
      cout << ctr;
    else
      cout << ctr + lt / 2;
    cout << " ";
    if (lt < 2 * (n - ctr))
      lt++;
    else {
      ctr++;
      lt = 1;
    }
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```