For a tree with up to a degree 2n-2
for every point we believe it to a degree
so for the rest of one weight corresponding to each
run can be completely determined when the size of the backpack n-2 is made the maximum value (initial negative infinity)
the size of the remaining equivalent of 1 to n-1, the capacity can be surely filled with n-2
#include<bits/stdc++.h>
#define inf 0x3f3f3f3f
#define mem(a,b) memset(a,b,sizeof(a))
#define sd(x) scanf("%d",&(x))
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
int dp[2050],w[2050];
int main()
{
int t;
sd(t);
while(t--)
{
int n;
sd(n);
rep(i,1,n) dp[i]=-inf;
dp[0]=0;
rep(i,1,n-1) sd(w[i]);
int sum=n*w[1];
rep(i,2,n) rep(j,i-1,n-2)
{
if(dp[j]<dp[j-i+1]+w[i]-w[1])
{
dp[j]=dp[j-i+1]+w[i]-w[1];
}
}
cout<<sum+dp[n-2]<<"\n";
}
return 0;
}