#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << endl; }
void clock_out() {
  cerr << "\nTime Elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " s\n";
}
void fileio() {
  freopen("/home/dwai/Desktop/cp/input.txt", "r", stdin);
  freopen("/home/dwai/Desktop/cp/output.txt", "w", stdout);
  freopen("/home/dwai/Desktop/cp/debug.txt", "w", stderr);
}
void ofileio() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
int IT_MAX = 1 << 20;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long lev = 1;
  long long cnt = 0;
  vector<long long> ans;
  while (cnt < r && lev < n) {
    long long x = (n - lev) * 2;
    if (cnt + x < l) {
      lev++;
      cnt += x;
      continue;
    }
    x = lev;
    while (cnt < r && x < n) {
      cnt++;
      if (cnt >= l && cnt <= r) {
        ans.emplace_back(lev);
      }
      cnt++;
      x++;
      if (cnt >= l && cnt <= r) {
        ans.emplace_back(x);
      }
    }
    lev++;
  }
  if (++cnt <= r) {
    ans.emplace_back(1);
  }
  for (long long x : ans) cout << x << " ";
  cout << "\n";
}
int main() {
  42;
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  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>
using namespace std;
void debug_out() { cerr << endl; }
void clock_out() {
  cerr << "\nTime Elapsed : " << 1.0 * clock() / CLOCKS_PER_SEC << " s\n";
}
void fileio() {
  freopen("/home/dwai/Desktop/cp/input.txt", "r", stdin);
  freopen("/home/dwai/Desktop/cp/output.txt", "w", stdout);
  freopen("/home/dwai/Desktop/cp/debug.txt", "w", stderr);
}
void ofileio() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << H;
  debug_out(T...);
}
int IT_MAX = 1 << 20;
const long long MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const long long LL_INF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
const double ERR = 1e-10;
void solve() {
  long long n, l, r;
  cin >> n >> l >> r;
  long long lev = 1;
  long long cnt = 0;
  vector<long long> ans;
  while (cnt < r && lev < n) {
    long long x = (n - lev) * 2;
    if (cnt + x < l) {
      lev++;
      cnt += x;
      continue;
    }
    x = lev;
    while (cnt < r && x < n) {
      cnt++;
      if (cnt >= l && cnt <= r) {
        ans.emplace_back(lev);
      }
      cnt++;
      x++;
      if (cnt >= l && cnt <= r) {
        ans.emplace_back(x);
      }
    }
    lev++;
  }
  if (++cnt <= r) {
    ans.emplace_back(1);
  }
  for (long long x : ans) cout << x << " ";
  cout << "\n";
}
int main() {
  42;
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  srand(time(NULL));
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```