#include <bits/stdc++.h>
#pragma GCC optimize "trapv"
const int inf1 = 1e9;
const long long int inf2 = 1e18;
const int N = 100000;
using namespace std;
int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int solve() {
  long long int n, l, r;
  cin >> n >> l >> r;
  if (l == n * (n - 1) + 1) {
    cout << 1 << "\n";
    return 0;
  }
  long long int bl = 1;
  bool f = false;
  if (r == (n * (n - 1) + 1)) {
    r--;
    f = true;
  }
  long long int lo = 1;
  long long int hi = n - 1;
  while (lo <= hi) {
    long long int m = (lo + hi) / 2;
    long long int ch = 2 * n * m - m * (m + 1);
    if (ch >= l) {
      bl = m;
      hi = m - 1;
    } else
      lo = m + 1;
  }
  long long int val = 2 * n * (bl - 1) - bl * (bl - 1);
  for (long long int i = l; i <= r; i++) {
    if (i % 2 == 0) {
      cout << bl + (i - val) / 2 << " ";
    } else {
      cout << bl << " ";
    }
    if ((2 * n * bl - (bl + 1) * bl) == i) {
      val = i;
      bl++;
    }
  }
  if (f) cout << 1 << " ";
  cout << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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>
#pragma GCC optimize "trapv"
const int inf1 = 1e9;
const long long int inf2 = 1e18;
const int N = 100000;
using namespace std;
int dirs[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
int solve() {
  long long int n, l, r;
  cin >> n >> l >> r;
  if (l == n * (n - 1) + 1) {
    cout << 1 << "\n";
    return 0;
  }
  long long int bl = 1;
  bool f = false;
  if (r == (n * (n - 1) + 1)) {
    r--;
    f = true;
  }
  long long int lo = 1;
  long long int hi = n - 1;
  while (lo <= hi) {
    long long int m = (lo + hi) / 2;
    long long int ch = 2 * n * m - m * (m + 1);
    if (ch >= l) {
      bl = m;
      hi = m - 1;
    } else
      lo = m + 1;
  }
  long long int val = 2 * n * (bl - 1) - bl * (bl - 1);
  for (long long int i = l; i <= r; i++) {
    if (i % 2 == 0) {
      cout << bl + (i - val) / 2 << " ";
    } else {
      cout << bl << " ";
    }
    if ((2 * n * bl - (bl + 1) * bl) == i) {
      val = i;
      bl++;
    }
  }
  if (f) cout << 1 << " ";
  cout << "\n";
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```