#include <iostream>
#include <cstring>
#include <vector>
using namespace std;const int MAX = 6;struct Tree{double L, R;    //树的左边和右边的长度Tree():L(0), R(0){}
};double width, w[MAX], sum[1<<MAX];
int vis[1<<MAX];
vector<Tree> tree[1<<MAX];        //枚举出来的结点可以组合成多个树void dfs(int subset);        //枚举二叉树int main(){freopen("input.txt", "r", stdin);int T;cin >> T;while(T--){int s;scanf("%lf%d", &width, &s);for(int i=0; i<s; i++){scanf("%lf", &w[i]);}//初始化 memset(sum, 0, sizeof(sum));for(int i=1; i<(1<<s); i++){tree[i].clear();    //清空树int k = 1;for(int j=0; j<s; j++){    //算出当前集合所有结点的重量和if(k & i){sum[i] += w[j];}k = (k << 1);} }//进行枚举int root = (1<<s) - 1;memset(vis, 0, sizeof(vis));dfs(root);//得到最大的答案double ans = -1;for(int i=0; i<tree[root].size(); i++){ans = max(tree[root][i].L + tree[root][i].R, ans);} printf("%.10lf\n", ans);}} void dfs(int subset){if(vis[subset])return;vis[subset] = 1;bool hasChild = false;for(int left = ((subset-1) & subset); left > 0; left = ((left-1) & subset)){    //枚举集合 hasChild = true;int right = subset ^ left;        //left 的补集double l = sum[right] / sum[subset];    //左臂长度double r = sum[left] / sum[subset];        //右臂长度
        dfs(left);    dfs(right);        //把左右树的子树也枚举出来for(int i=0; i<tree[left].size(); i++){            //将枚举出的左右子树组合起来 for(int j=0; j<tree[right].size(); j++){Tree t;t.L = max(tree[left][i].L + l, tree[right][j].L - r);t.R = max(tree[right][j].R + r, tree[left][i].R - l);if(t.L + t.R < width){                    //如果满足条件，枚举结果加一
                    tree[subset].push_back(t); }}}}if(!hasChild){        //叶子结点的左右臂为零
        tree[subset].push_back(Tree());}}