text: # 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.# Solution #include <bits/stdc++.h> using namespace std; long long n; int tc = 0; int solve(); int main() { ios_base::sync_with_stdio(0); if (tc < 0) { cout << "TC!\n"; cin.ignore(1e8); } else if (!