题目网址：

https://vjudge.net/problem/HDU-6092

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them: 

Yuta has nn positive A1−AnA1−An and their sum is mm. Then for each subset SS of AA, Yuta calculates the sum of SS. 

Now, Yuta has got 2n2n numbers between [0,m][0,m]. For each i∈[0,m]i∈[0,m], he counts the number of iis he got as BiBi. 

Yuta shows Rikka the array BiBi and he wants Rikka to restore A1−AnA1−An. 

It is too difficult for Rikka. Can you help her?  

Input

The first line contains a number t(1≤t≤70)t(1≤t≤70), the number of the testcases. 

For each testcase, the first line contains two numbers n,m(1≤n≤50,1≤m≤104)n,m(1≤n≤50,1≤m≤104).

The second line contains m+1m+1 numbers B0−Bm(0≤Bi≤2n)B0−Bm(0≤Bi≤2n).

Output

For each testcase, print a single line with nn numbers A1−AnA1−An. 

It is guaranteed that there exists at least one solution. And if there are different solutions, print the lexicographic minimum one.

Sample Input

2
2 3
1 1 1 1
3 3
1 3 3 1

Sample Output

1 2
1 1 1 

Hint

In the first sample, $A$ is $[1,2]$. $A$ has four subsets $[],[1],[2],[1,2]$ and the sums of each subset are $0,1,2,3$. So $B=[1,1,1,1]$

解题思路：

用dp[j]来表示组成 j 所需的方案数，那么对于数 i 来说，数 i 的个数就是子序列中和为 i 的个数即 b[i] 减去其他数字组成 i 的个数即 dp[i]，由此可以算出数字 i 的个数。

由于dp[j]来表示组成 j 所需的方案数，那么对于dp[j]来说，dp[j] = dp[j] + dp[j-i]

因此我们可以先遍历要确定的数字即最外层的循环 for(ll i=1;i<=m;i++)，再确定数字 i 的个数，内两层循环相当于一个01背包，以确定更新dp数组

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
ll a[120];
ll dp[10100];
ll b[10020];
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);int t;scanf("%d",&t);while(t--) {ms(dp);int n,m;scanf("%d %d",&n,&m);rep(i,0,m) scanf("%I64d",&b[i]);dp[0]=1;int cnt=0;for(ll i=1;i<=m;i++) {if(b[i]) {if(cnt==n) break;ll num=b[i]-dp[i];for(ll j=1;j<=num;j++) {a[++cnt]=i;for(ll k=m;k>=i;k--) {dp[k]=dp[k]+dp[k-i];}}}}for(int i=1;i<cnt;i++) {printf("%I64d ",a[i]);}printf("%I64d\n",a[cnt]);}return 0;
}