CF1110E.Magic Stones

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 $i$?

格里戈里有 $n$ 颗魔法石子，编号从 $1$ 到 $n$。第 $i$ 颗石子的电荷量为 $c_i$。

有时格里戈里感到无聊，便会选取某颗内部石子（即下标 $i$ 满足 $2 \le i \le n - 1$ 的石子），并使其与相邻石子同步。同步后，该石子失去自身原有的电荷，但获得相邻石子的电荷组合效果：其电荷 $c_i$ 变为 $c_i' = c_{i + 1} + c_{i - 1} - c_i$。

格里戈里的朋友安德鲁也有 $n$ 颗石子，电荷量分别为 $t_i$。格里戈里很好奇：是否存在一个由零次或多次同步操作组成的序列，能将他自己石子的电荷量 $c_i$ 全部变为安德鲁对应石子的电荷量 $t_i$（即对所有 $i$，均有 $c_i \to t_i$）？

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.

第一行包含一个整数 $n$（$2 \le n \le 10^5$）—— 魔法石的数量。

第二行包含整数 $c_1, c_2, \ldots, c_n$（$0 \le c_i \le 2 \cdot 10^9$）—— 格里戈里的石头的电荷量。

第三行包含整数 $t_1, t_2, \ldots, t_n$（$0 \le t_i \le 2 \cdot 10^9$）—— 安德鲁的石头的电荷量。

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

Otherwise, print "No".

如果存在一个（可能为空的）同步操作序列，能够将所有电荷变为所需值，则输出“Yes”。

否则，输出“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.

在第一个例子中，我们可以执行以下同步操作（从 1 开始编号）：

• 首先，同步第三颗石子：$[7, 2, \mathbf{4}, 12] \rightarrow [7, 2, \mathbf{10}, 12]$。
• 然后，同步第二颗石子：$[7, \mathbf{2}, 10, 12] \rightarrow [7, \mathbf{15}, 10, 12]$。

在第二个例子中，对第二颗石子执行任何操作都不会改变其电荷量。