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A12456. Magic Stones

编程题 普及/提高-

题目描述

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 $i$ ?

输入格式

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.

输出格式

If there exists a (possibly empty) sequence of synchronization operations, which changes all charges to the required ones, print "Yes".

Otherwise, print "No".

输入输出样例

输入 #1
4
7 2 4 12
7 15 10 12
输出 #1
Yes
输入 #2
3
4 4 4
1 2 3
输出 #2
No

说明/提示

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.
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