Fishing Master

Heard that  eom is a fishing MASTER, you want to acknowledge him as your mentor. As everybody knows, if you want to be a MASTER's apprentice, you should pass the trial. So when you find fishing MASTER eom, the trial is as follow:

There are n fish in the pool. For the i - ℎth fish, it takes at least ti​ minutes to stew(overcook is acceptable). To simplify this problem, the time spent catching a fish is k minutes. You can catch fish one at a time and because there is only one pot, only one fish can be stewed in the pot at a time. While you are catching a fish, you can not put a raw fish you have caught into the pot, that means if you begin to catch a fish, you can't stop until after k minutes; when you are not catching fish, you can take a cooked fish (stewed for no less than ti​) out of the pot or put a raw fish into the pot, these two operations take no time. Note that if the fish stewed in the pot is not stewed for enough time, you cannot take it out, but you can go to catch another fish or just wait for a while doing nothing until it is sufficiently stewed.

Now eom wants you to catch and stew all the fish as soon as possible (you definitely know that a fish can be eaten only after sufficiently stewed), so that he can have a satisfying meal. If you can complete that in the shortest possible time, eom will accept you as his apprentice and say "I am done! I am full!". If you can't, eom will not accept you and say "You are done! You are fool!".

So what's the shortest time to pass the trial if you arrange the time optimally?

 

Input

The first line of input consists of a single integer T(1≤T≤20), denoting the number of test cases.

For each test case, the first line contains two integers n(1≤n≤105),k(1≤k≤109), denoting the number of fish in the pool and the time needed to catch a fish.

the second line contains n integers,t1​,t2​,…,tn​(1≤ti​≤109) ,denoting the least time needed to cook the i - ℎth fish.

 

Output

For each test case, print a single integer in one line, denoting the shortest time to pass the trial.

Inputcopy	Outputcopy
2 3 5 5 5 8 2 4 3 3	23 11

 

Hint

Case 1: Catch the 3rd fish (5 mins), put the 3rd fish in, catch the 1st fish (5 mins), wait (3 mins),

take the 3rd fish out, put the 1st fish in, catch the 2nd fish(5 mins),

take the 1st fish out, put the 2nd fish in, wait (5 mins), take the 2nd fish out.

Case 2: Catch the 1st fish (4 mins), put the 1st fish in, catch the 2nd fish (4 mins),

take the 1st fish out, put the 2nd fish in, wait (3 mins), take the 2nd fish out.

代码思路：先求钓鱼和煮鱼的总和，然后算出煮鱼时可以钓鱼的数量，然后减去相应的时间（有可能煮鱼时间内能钓上的鱼大于给的鱼条数，需要判断），最后减去煮鱼干等着的时间就可以了。

#include<bits/stdc++.h>
using namespace std;

long long t,n,m,x,y,z,sum,a;
priority_queue<long long,vector<long long>, less<long long> > q;

int main(){
	scanf("%lld",&t);
	//t组样例 
	while(t--){
		while(!q.empty())q.pop();
		//n为鱼的数量，m为钓一条鱼的时间 
		scanf("%lld %lld",&n,&m);
		//x为煮鱼时间内可钓到鱼的条数，sum先算上所有钓鱼时间总长 
		x=0;sum=n*m;
		for(int i=0;i<n;i++){
			scanf("%lld",&a);
			//sum加上每一条鱼的煮鱼时间 
			sum+=a;
			x+=a/m;
			//压入煮鱼时干等着的时间 
			q.push(a%m);
		}
		//防止钓鱼时间变为负数且保证先钓后煮 
		if(x>=n-1)x=n-1;
		//减去煮鱼时钓鱼的时间 
		sum-=x*m;
		//算有多少次可以不干等着 
		x=n-1-x;
		while(x--){
			if(!q.empty()){
				//减去干等着的时间 
				sum-=q.top();
				q.pop();
			}else{
				break;
			}
		}
		printf("%lld\n",sum);
	}
	return 0;
}