Grigory has nn magic stones, conveniently numbered from 11 to nn. The charge of the ii-th stone is equal to cici.
Sometimes Grigory gets bored and selects some inner stone (that is, some stone with index ii, where 2≤i≤n−12≤i≤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 cici changes to c′i=ci+1+ci−1−cici′=ci+1+ci−1−ci.
Andrew, Grigory's friend, also has nn stones with charges titi. 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 cici into titi for all ii?
Input
The first line contains one integer nn (2≤n≤1052≤n≤105) — the number of magic stones.
The second line contains integers c1,c2,…,cnc1,c2,…,cn (0≤ci≤2⋅1090≤ci≤2⋅109) — the charges of Grigory's stones.
The second line contains integers t1,t2,…,tnt1,t2,…,tn (0≤ti≤2⋅1090≤ti≤2⋅109) — the charges of Andrew's stones.
Output
If there exists a (possibly empty) sequence of synchronization operations, which changes all charges to the required ones, print "Yes".
Otherwise, print "No".
Examples
input
Copy
4 7 2 4 12 7 15 10 12
output
Copy
Yes
input
Copy
3 4 4 4 1 2 3
output
Copy
No
Note
In the first example, we can perform the following synchronizations (11-indexed):
- First, synchronize the third stone [7,2,4,12]→[7,2,10,12][7,2,4,12]→[7,2,10,12].
- Then synchronize the second stone: [7,2,10,12]→[7,15,10,12][7,2,10,12]→[7,15,10,12].
In the second example, any operation with the second stone will not change its charge.
题意：
就是按题目的操作能不能把两个数组变成一样。
思路：写几个例子可以发现最终都是他们的差分换来换去，所以只要判断它们的差分是否相同即可。（要注意最后一项和第一项要判断一下值是否相同因为他们是不变的）
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define LL long long
using namespace std;
const int maxn=1e5+100;
int a[maxn],b[maxn];
int c1[maxn],c2[maxn];
int main()
{
int n;
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
for(int i=1;i<=n;i++)
{
scanf("%d",&b[i]);
}
if(a[1]!=b[1])
{
printf("NO\n");
return 0;
}
c1[n]=a[n];
for(int i=1;i<n;i++)
{
c1[i]=a[i+1]-a[i];
}
c2[n]=a[n];
for(int i=1;i<n;i++)
{
c2[i]=b[i+1]-b[i];
}
sort(c1+1,c1+n+1);
sort(c2+1,c2+1+n);
for(int i=1;i<=n;i++)
{
if(c1[i]!=c2[i])
{
printf("NO\n");
return 0;
}
}
printf("YES\n");
return 0;
}