Magic Stones

**题意：** 一次操作选择$1<i<n$，使$c_i$变为$c_i'$，$c_i'=c_{i+1}+c_{i-1}-c_i$ 是否能做若干次操作，使每个$c_i$和$t_i$相等？ **数据范围：** $2\le n\le 10^5, 0\le c_i\le 2*10^9, 0\le t_i\le 2*10^9$ **输入方式：** 行1 $n$ 行2 $c_1, \cdots, c_n$ 行3 $t_1, \cdots, t_n$ **输出方式：** Yes/No

Grigory has $ n $ magic stones, conveniently numbered from $ 1 $ to $ n $ . The charge of the $ i $ -th stone is equal to $ c_i $ . Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index $ i $ , where $ 2 \le i \le n - 1 $ ), and after that synchronizes it with neighboring stones. After that, the chosen stone loses its own charge, but acquires the charges from neighboring stones. In other words, its charge $ c_i $ changes to $ c_i' = c_{i + 1} + c_{i - 1} - c_i $ . Andrew, Grigory's friend, also has $ n $ stones with charges $ t_i $ . Grigory is curious, whether there exists a sequence of zero or more synchronization operations, which transforms charges of Grigory's stones into charges of corresponding Andrew's stones, that is, changes $ c_i $ into $ t_i $ for all $ i $ ?

The first line contains one integer $ n $ ( $ 2 \le n \le 10^5 $ ) — the number of magic stones. The second line contains integers $ c_1, c_2, \ldots, c_n $ ( $ 0 \le c_i \le 2 \cdot 10^9 $ ) — the charges of Grigory's stones. The second line contains integers $ t_1, t_2, \ldots, t_n $ ( $ 0 \le t_i \le 2 \cdot 10^9 $ ) — the charges of Andrew's stones.

If there exists a (possibly empty) sequence of synchronization operations, which changes all charges to the required ones, print "Yes". Otherwise, print "No".

4
7 2 4 12
7 15 10 12

Yes

3
4 4 4
1 2 3

No

In the first example, we can perform the following synchronizations ( $ 1 $ -indexed): - First, synchronize the third stone $ [7, 2, \mathbf{4}, 12] \rightarrow [7, 2, \mathbf{10}, 12] $ . - Then synchronize the second stone: $ [7, \mathbf{2}, 10, 12] \rightarrow [7, \mathbf{15}, 10, 12] $ . In the second example, any operation with the second stone will not change its charge.