#include <bits/stdc++.h>
#pragma GCC optimize("O3", "unroll-all-loops")
using namespace std;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
long long n, l, r;
bool viv = false;
vector<long long> v;
void solve() {
  cin >> n >> l >> r;
  l--, r--;
  vector<pair<long long, long long> > v;
  for (int i = 0; i < n; i++) {
    v.push_back({i, 2 * (n - 1 - i)});
  }
  long long sl = 0;
  vector<long long> ans;
  for (int i = 0; i < n; i++) {
    long long sr = sl + v[i].second - 1;
    if (max(l, sl) <= min(r, sr)) {
      vector<long long> loc;
      for (int j = 0; j < v[i].second / 2; j++) {
        loc.push_back(i);
        loc.push_back(i + j + 1);
      }
      for (int j = 0; j < loc.size(); j++)
        if (sl + j >= l && sl + j <= r) ans.push_back(loc[j]);
    }
    sl = sr + 1;
  }
  if (r == sl) ans.push_back(0);
  for (auto i : ans) cout << i + 1 << ' ';
  cout << '\n';
}
int main() {
  viv = true;
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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", "unroll-all-loops")
using namespace std;
ifstream in;
ofstream out;
const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = ml * kk + 7;
const long long inf = ml * ml * ml + 7;
long long n, l, r;
bool viv = false;
vector<long long> v;
void solve() {
  cin >> n >> l >> r;
  l--, r--;
  vector<pair<long long, long long> > v;
  for (int i = 0; i < n; i++) {
    v.push_back({i, 2 * (n - 1 - i)});
  }
  long long sl = 0;
  vector<long long> ans;
  for (int i = 0; i < n; i++) {
    long long sr = sl + v[i].second - 1;
    if (max(l, sl) <= min(r, sr)) {
      vector<long long> loc;
      for (int j = 0; j < v[i].second / 2; j++) {
        loc.push_back(i);
        loc.push_back(i + j + 1);
      }
      for (int j = 0; j < loc.size(); j++)
        if (sl + j >= l && sl + j <= r) ans.push_back(loc[j]);
    }
    sl = sr + 1;
  }
  if (r == sl) ans.push_back(0);
  for (auto i : ans) cout << i + 1 << ' ';
  cout << '\n';
}
int main() {
  viv = true;
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```