Codeforces Global Round 1
E
Magic Stones
Codeforces
2200
原题
← DF →
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Grigory has $n$ magic stones, conveniently numbered from $1$ to $n$. The charge of the $i$-th stone is equal to $c_i$.

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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 $c_i' = c_{i + 1} + c_{i - 1} - c_i$.

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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 $i$?

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Input

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The first line contains one integer $n$ ($2 \le n \le 10^5$) — the number of magic stones.

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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.

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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.

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Output

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

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Otherwise, print "No".

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Note

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In the first example, we can perform the following synchronizations ($1$-indexed):

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  • First, synchronize the third stone $[7, 2, \mathbf{4}, 12] \rightarrow [7, 2, \mathbf{10}, 12]$.
  • Then synchronize the second stone: $[7, \mathbf{2}, 10, 12] \rightarrow [7, \mathbf{15}, 10, 12]$.
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In the second example, any operation with the second stone will not change its charge.

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样例

Input #1
4
7 2 4 12
7 15 10 12
Output #1
Yes
InputCopy OutputCopy
Input #2
3
4 4 4
1 2 3
Output #2
No
InputCopy OutputCopy
题目信息
题号
1110E
时间限制
1 s
内存限制
256 MB
标签
构造
数学
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