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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 nn 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 n−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 ff is a predefined function and dd 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

232 145 1 4

Sample Output

519

题意：构造一棵树，每个点的度数为 ai，求 f(a1) + ... + f(an) 的最大值

思路：一棵n个点的树的总度数为2 * n - 2，先给每个点都分一度（肯定的），剩下的 n - 2 个度作为背包容量，度数 0 ~ n - 2 看作 n - 1 个物品，就变成了一个裸的完全背包

#include 
   
    using namespace std;typedef long long ll;const int inf = 0x3f3f3f3f;const int N = 2020;int dp[2 * N], f[N];int main() {    int t, n;    scanf("%d", &t);    while(t--) {        scanf("%d", &n);        for(int i = 1; i < n; ++i) scanf("%d", &f[i]);        for(int i = 0; i < N; ++i) dp[i] = -inf;        dp[0] = 0;        for(int i = 1; i < n - 1; ++i) {            for(int j = i; j <= n - 2; ++j) {                dp[j] = max(dp[j], dp[j - i] + f[i + 1] - f[1]);            }        }        printf("%d\n", dp[n - 2] + n * f[1]);    }	return 0;}