Industrial Nim

第一行给出 $N$ 表示有 $N$ 个采石场，接下来 $N$ 行每一行一个 $X_i$ 一个 $M_i$，表示第 $i$ 个采石场有 $M_i$ 辆车，这个采石场中第 $1$ 辆车的石头数量是 $X_i$，第 $2$ 辆车中有 $X_i+1$ 个石头......第 $M_i$ 辆车的石头的数量是 $X_i+M_i-1$。 有两个人每次从任意一辆车中取任意数量的石头（不能为 $0$ ），最后一个取完所有石头的赢。若先手赢输出"tolik"，后手赢输出"bolik"。 $1\le n\le 10^5$ $1\le x_i,m_i\le 10^{16}$

There are $ n $ stone quarries in Petrograd. Each quarry owns $ m_{i} $ dumpers ( $ 1<=i<=n $ ). It is known that the first dumper of the $ i $ -th quarry has $ x_{i} $ stones in it, the second dumper has $ x_{i}+1 $ stones in it, the third has $ x_{i}+2 $ , and the $ m_{i} $ -th dumper (the last for the $ i $ -th quarry) has $ x_{i}+m_{i}-1 $ stones in it. Two oligarchs play a well-known game Nim. Players take turns removing stones from dumpers. On each turn, a player can select any dumper and remove any non-zero amount of stones from it. The player who cannot take a stone loses. Your task is to find out which oligarch will win, provided that both of them play optimally. The oligarchs asked you not to reveal their names. So, let's call the one who takes the first stone «tolik» and the other one «bolik».

The first line of the input contains one integer number $ n $ ( $ 1<=n<=10^{5} $ ) — the amount of quarries. Then there follow $ n $ lines, each of them contains two space-separated integers $ x_{i} $ and $ m_{i} $ ( $ 1<=x_{i},m_{i}<=10^{16} $ ) — the amount of stones in the first dumper of the $ i $ -th quarry and the number of dumpers at the $ i $ -th quarry.

Output «tolik» if the oligarch who takes a stone first wins, and «bolik» otherwise.

2
2 1
3 2

tolik

4
1 1
1 1
1 1
1 1

bolik