#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
using namespace std;
const double PI = acos(-1);
long long gcd() { return 0ll; }
template <typename T, typename... Args>
T gcd(T a, Args... args) {
  return __gcd(a, (__typeof(a))gcd(args...));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    long long n, l, r;
    cin >> n >> l >> r;
    l--;
    r--;
    long long a[2 * n - 3];
    for (__typeof(2 * n - 3) i = (0) - ((0) > (2 * n - 3));
         i != (2 * n - 3) - ((0) > (2 * n - 3));
         i += 1 - 2 * ((0) > (2 * n - 3))) {
      if (i & 1)
        a[i] = 1 + ((i + 1) / 2);
      else
        a[i] = 1;
    }
    while (l < (2 * n - 3) && l <= r) cout << a[l++] << " ";
    long long st[n - 1], x = 2;
    st[n - 2] = n * (n - 1) - 1;
    for (__typeof(0) i = (n - 2) - ((n - 2) > (0)); i != (0) - ((n - 2) > (0));
         i += 1 - 2 * ((n - 2) > (0))) {
      st[i] = st[i + 1] - x;
      x += 2;
    }
    long long j = -1;
    while (st[j + 1] < l) j++;
    while (l <= r) {
      if (l == (n * (n - 1))) {
        cout << "1";
        break;
      }
      if (st[j + 1] == l) {
        j++;
        l++;
        cout << n << " ";
        continue;
      }
      if (!(l & 1))
        cout << (j + 2) << " ";
      else
        cout << (j + 2) + (l - st[j]) / 2 << " ";
      l++;
    }
    cout << "\n";
  }
  return 0;
}

### 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>
#pragma GCC optimize("-O3")
using namespace std;
const double PI = acos(-1);
long long gcd() { return 0ll; }
template <typename T, typename... Args>
T gcd(T a, Args... args) {
  return __gcd(a, (__typeof(a))gcd(args...));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int T;
  cin >> T;
  while (T--) {
    long long n, l, r;
    cin >> n >> l >> r;
    l--;
    r--;
    long long a[2 * n - 3];
    for (__typeof(2 * n - 3) i = (0) - ((0) > (2 * n - 3));
         i != (2 * n - 3) - ((0) > (2 * n - 3));
         i += 1 - 2 * ((0) > (2 * n - 3))) {
      if (i & 1)
        a[i] = 1 + ((i + 1) / 2);
      else
        a[i] = 1;
    }
    while (l < (2 * n - 3) && l <= r) cout << a[l++] << " ";
    long long st[n - 1], x = 2;
    st[n - 2] = n * (n - 1) - 1;
    for (__typeof(0) i = (n - 2) - ((n - 2) > (0)); i != (0) - ((n - 2) > (0));
         i += 1 - 2 * ((n - 2) > (0))) {
      st[i] = st[i + 1] - x;
      x += 2;
    }
    long long j = -1;
    while (st[j + 1] < l) j++;
    while (l <= r) {
      if (l == (n * (n - 1))) {
        cout << "1";
        break;
      }
      if (st[j + 1] == l) {
        j++;
        l++;
        cout << n << " ";
        continue;
      }
      if (!(l & 1))
        cout << (j + 2) << " ";
      else
        cout << (j + 2) + (l - st[j]) / 2 << " ";
      l++;
    }
    cout << "\n";
  }
  return 0;
}

```