You are given two integer arrays a and b of length n.

You can reverse at most one subarray (continuous subsegment) of the array a. 

Your task is to reverse such a subarray that the sum ∑_{i=1}^n a_i ⋅ b_i is maximized.

Input

The first line contains one integer n (1 ≤ n ≤ 5000).

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^7).

The third line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^7).

Output

Print single integer — maximum possible sum after reversing at most one subarray (continuous subsegment) of a.

Examples

Input


5
2 3 2 1 3
1 3 2 4 2


Output


29


Input


2
13 37
2 4


Output


174


Input


6
1 8 7 6 3 6
5 9 6 8 8 6


Output


235

Note

In the first example, you can reverse the subarray [4, 5]. Then a = [2, 3, 2, 3, 1] and 2 ⋅ 1 + 3 ⋅ 3 + 2 ⋅ 2 + 3 ⋅ 4 + 1 ⋅ 2 = 29.

In the second example, you don't need to use the reverse operation. 13 ⋅ 2 + 37 ⋅ 4 = 174.

In the third example, you can reverse the subarray [3, 5]. Then a = [1, 8, 3, 6, 7, 6] and 1 ⋅ 5 + 8 ⋅ 9 + 3 ⋅ 6 + 6 ⋅ 8 + 7 ⋅ 8 + 6 ⋅ 6 = 235.
#include<iostream>
#include<stdio.h>
#include<cmath>
#include<stdlib.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<vector>
#include<map>
#define oo 0x3f3f3f3f
#define OO 0x3f3f3f3f3f3f3f3f
#define LL long long 
#define res register int
#define STP system("pause")
using namespace std;
const int N=5e3;
LL f[N+5][N+5];
int a[N+5],b[N+5];
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&a[i]);
		f[i][i]=0;
	}
	for(int i=1;i<=n;++i){
		scanf("%d",&b[i]);
	}
	LL ans=0;
	for(int k=2;k<=n;++k){
		for(int i=1;i<=n-k+1;++i){
			f[i][i+k-1]=f[i+1][i+k-2]+(LL)(a[i]-a[i+k-1])*(b[i]-b[i+k-1]);
			ans=min(ans,f[i][i+k-1]);
		}
	}
//	cout<<ans<<'\n';
//	STP;
	for(int i=1;i<=n;++i)
		ans-=(LL)a[i]*b[i];
	cout<<-ans<<'\n';
	return 0;
}