Red packet

Year is coming! Our big boss Wine93 will distribute some “Red Package”, just like Alipay and Wechat.

Wine93 has m yuan, he decides to distribute them to n people and everyone can get some money(0 yuan is not allowed and everyone’s money is an integer), Now k people has gotten money, it’s your turn to get “Red Package”, you want to know, at least how much money to give you, then you can must become the “lucky man”. and the m yuan must be used out.

Noting that if someone’s money is strictly much than others’, than he is “lucky man”.

Input

Input starts with an integer T (T <= 50) denoting the number of test case. 
For each test case, three integers n, m, k (1 <= k < n <= 100000, 0< m <= 100000000) will be given. 
Next line contains k integers, denoting the money that k people get. You can assume that the k integers’ summation is no more than m. 

Output

Ouput the least money that you need to become the “lucky man”, if it is impossible, output “Impossible” (no quote). 

Sample Input

3
3 5 2
2 1
4 10 2
2 3
4 15 2
3 5

Sample Output

Impossible
4

6

#include<cstdio>//这个题用int就可以过  我之前用long  long一直错  可能是该题的网站不支持long long 做题时要注意 
#include<cstring>
int  money[100001];
int main()
{
	int t;
	scanf("%d",&t); 
	while(t--)
	{
		int  n,m,k;
		scanf("%d%d%d",&n,&m,&k);
		int  sum=0; 
		int  max=0;
		for(int i=0;i<k;i++)
		{
			scanf("%d",&money[i]);
			sum+=money[i];
			if(max<money[i])
			{
				max=money[i];
			}
		 } 
		 int  qian=m-sum;
		 int  ren=n-k;	
          int  q=qian-ren+1;
		 
		if(q<=max)
		{
				printf("Impossible\n");
		}
		else
		{ 
		int  l=max+1 , r=q;
			int  mid;
			while(l<=r)
			{
				mid=(l+r)/2;
				if(mid<=q-mid+1)
				{
					l=mid+1;
				}
				else
				{
					r=mid-1;
				}
			}
				printf("%d\n",l);
		}
	}
	return 0;
}