A prince of the Science Continent was imprisoned in a castle because of his contempt for mathematics when he was young, and was entangled in some mathematical curses. He studied hard until he reached adulthood and decided to use his knowledge to escape the castle.

There are NN rooms from the place where he was imprisoned to the exit of the castle. In the i^{th}ith room, there is a wizard who has a resentment value of a[i]a[i]. The prince has MM curses, the j^{th}jth curse is f[j]f[j], and f[j]f[j] represents one of the four arithmetic operations, namely addition('+'

), subtraction('-'), multiplication('*'), and integer division('/'). The prince's initial resentment value is KK. Entering a room and fighting with the wizard will eliminate a curse, but the prince's resentment value will become the result of the arithmetic operation f[j]f[j] with the wizard's resentment value. That is, if the prince eliminates the j^{th}jth curse in the i^{th}ith room, then his resentment value will change from xx to (x\ f[j]\ a[i]x f[j] a[i]), for example, when x=1, a[i]=2, f[j]=x=1,a[i]=2,f[j]='+', then xx will become 1+2=31+2=3.

Before the prince escapes from the castle, he must eliminate all the curses. He must go from a[1]a[1] to a[N]a[N] in order and cannot turn back. He must also eliminate the f[1]f[1] to f[M]f[M] curses in order(It is guaranteed that N\ge MN≥M). What is the maximum resentment value that the prince may have when he leaves the castle?

Input

The first line contains an integer T(1 \le T \le 1000)T(1≤T≤1000), which is the number of test cases.

For each test case, the first line contains three non-zero integers: N(1 \le N \le 1000), M(1 \le M \le 5)N(1≤N≤1000),M(1≤M≤5) and K(-1000 \le K \le 1000K(−1000≤K≤1000), the second line contains NN non-zero integers: a[1], a[2], ..., a[N](-1000 \le a[i] \le 1000)a[1],a[2],...,a[N](−1000≤a[i]≤1000), and the third line contains MM characters: f[1], f[2], ..., f[M](f[j] =f[1],f[2],...,f[M](f[j]='+','-','*','/', with no spaces in between.

Output

For each test case, output one line containing a single integer.

樣例輸入複製

3
2 1 5
2 3
/
3 2 1
1 2 3
++
4 4 5
1 2 3 4
+-*/

樣例輸出複製

2
6
3

題目來源

題意：有很多個房間，房間裡有一個巫師，能量值為a，王子有4種操作，跟巫師戰鬥會讓他的能量值變化，問王子用完所有操作之後的最大值。操作必須按順序用，房間也必須按順序打。但是可以選擇不打。

解題思路：赤裸裸的DP了，開場題意看了好久……dp[i][j]表示打了前i個房間，用了j種操作的最大值和最小值。然後遍歷的時候就分兩種情況，打或不打。直接遞推過去就好了。重點是要記錄最大值和最小值，因為有負數的存在。這裡直接過載運算子，很方便的一個操作。

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
const long long INF=1e15+5;
const int maxn=1e4+5;
struct node{
	long long max,min;
	node operator +(const int &a) const { return {max+a,min+a};};
	node operator -(const int &a) const { return {max-a,min-a};};
	node operator *(const int &a) const {
		long long arr[2]={max*a,min*a};
		sort(arr,arr+2);
		return {arr[1],arr[0]};
	}
	node operator /(const int &a) const {
		long long arr[5]={max/a,min/a};
		sort(arr,arr+2);
		return {arr[1],arr[0]};
	}
}dp[maxn][10];
int n,m,k;
long long a[maxn];
char s[10];
int main()
{
	int T;
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d%d",&n,&m,&k);
		for (int i=1;i<=n;i++)
			scanf("%lld",&a[i]);
		scanf("%s",s);
		for (int i=1;i<=m;i++)
			dp[0][i].max=-INF,dp[0][i].min=INF;
		dp[0][0].max=dp[0][0].min=k;
		node u;
		for (int i=1;i<=n;i++)
		{
			for (int j=0;j<=m;j++) dp[i][j]=dp[i-1][j];//不打的情況
			for (int j=1;j<=m;j++)//打的情況，直接遞推
			{
				if (j>i) break;
				switch (s[j-1]) 
				{
					case '+':{u=dp[i-1][j-1]+a[i];break;}
					case '-':{u=dp[i-1][j-1]-a[i];break;}
					case '*':{u=dp[i-1][j-1]*a[i];break;}
					case '/':{u=dp[i-1][j-1]/a[i];break;}
				}
				dp[i][j].max=max(dp[i][j].max,u.max);
				dp[i][j].min=min(dp[i][j].min,u.min);
			}
		}
		printf("%lld\n",dp[n][m].max);
	}
}