In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly one path. In other words, any connected graph without simple cycles is a tree. You find a partial tree on the way home. This tree has n nodes but lacks of n−1 edges. You want to complete this tree by adding n−1 edges. There must be exactly one path between any two nodes after adding. As you know, there are nn−2 ways to complete this tree, and you want to make the completed tree as cool as possible. The coolness of a tree is the sum of coolness of its nodes. The coolness of a node is f(d), where f is a predefined function and d

is the degree of this node. What's the maximum coolness of the completed tree?

Input

The first line contains an integer T

indicating the total number of test cases. Each test case starts with an integer n in one line, then one line with n−1 integers f(1),f(2),…,f(n−1). 1≤T≤2015 2≤n≤2015 0≤f(i)≤10000 There are at most 10 test cases with n>100

.

Output

For each test case, please output the maximum coolness of the completed tree in one line.

Sample Input

2
3
2 1
4
5 1 4

Sample Output

5
19

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<iostream>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}
#define MS(x,y) memset(x,y,sizeof(x))
#define MC(x,y) memcpy(x,y,sizeof(x))
#define MP(x,y) make_pair(x,y)
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}
template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}
const int N=0,M=0,Z=1e9+7,ms63=1061109567;
int casenum,casei;
int n;
int a[2020],d[2020];
int f[2020];
int main()
{
	scanf("%d",&casenum);
	for(casei=1;casei<=casenum;casei++)
	{
		scanf("%d",&n);
		for(int i=1;i<n;i++)scanf("%d",&a[i]);
		for(int i=1;i<n;i++)d[i-1]=a[i]-a[1];
		MS(f,0);f[0]=n*a[1];
		int top=n-2;
		for(int i=1;i<=top;i++)
		{
			for(int j=1;j<=i;j++)gmax(f[i],f[i-j]+d[j]);
		}
		printf("%d\n",f[top]);
	}
	return 0;
}