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

題解：當到達一個個房間時在此房間做運算操作可以得到一個大值和一個小值如果運算子中存在乘除運算並且在逐個運算的個過程中遇到負數且為乘除操作此時就需要用前面的得到的最小值來進行運算操作，所有我們只需要維護達到第i個房間時用第j種運算子時得到的最大值和最小值最後返回dpmax[n][m[就好。

#include<cstring>
#include<cstdio>
#include<iostream>
#include<string>
#include<queue>
#include<vector>
#include<algorithm>
#include<cmath>
#include<set>
#include<map>
#include<stack>
#include<functional>
using namespace std;
#define clr(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&-x
#define rep(a,b,c) for(ll a=b;a<c;a++)
#define dec(a,b,c) for(int a=b;a>c;a--)
#define eb(x) emplace_back(x)
#define pb(x) push_back(x)
#define ps(x) push(x)
#define MAX_N 100000+5
#define MAX_M 100
typedef long long ll;
typedef unsigned long long ull;
typedef priority_queue<ll,vector<ll>,greater<ll> >pqg;
const ll maxn=1000+5;
const ll inf=1e7;
ll mod=1e9+7;
int n,m,k;
ll room[maxn];
ll dpmax[maxn][6],dpmin[maxn][6];
char f[7];
ll solve()
{
    ll ans=-inf;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m&&j<=i;j++)
        {
            ll tmax=dpmax[i-1][j-1],tmin=dpmin[i-1][j-1];
            ll maxx,minn;
            switch(f[j])
            {
            case '+':
                {
                    maxx=tmax+room[i];
                    minn=tmin+room[i];
                    break;
                }
            case '-':
                {
                    maxx=tmax-room[i];
                    minn=tmin-room[i];
                    break;
                }
            case '*':
                {
                    maxx=max(tmax*room[i],tmin*room[i]);
                    minn=min(tmin*room[i],tmax*room[i]);
                    break;
                }
            case '/':
                {
                    maxx=max(tmax/room[i],tmin/room[i]);
                    minn=min(tmin/room[i],tmax/room[i]);
                    break;
                }
            }
            dpmax[i][j]=max(dpmax[i-1][j],maxx);
            dpmin[i][j]=min(dpmin[i-1][j],minn);
            if(i==j){dpmax[i][j]=maxx;dpmin[i][j]=minn;}
        }
    }
    return dpmax[n][m];
}
int main()
{
#ifndef ONLINE_JUDGE
    // freopen("data.txt","r",stdin);
#endif
    int t;
    scanf("%d",&t);
    while(t--)
    {
      clr(dpmax,0);
      clr(dpmin,0);
      scanf("%d%d%d",&n,&m,&k);
      for(int i=1;i<=n;i++)
      {
          scanf("%lld",&room[i]);
          dpmin[i][0]=dpmax[i][0]=k;
      }
      dpmin[0][0]=dpmax[0][0]=k;
      cin>>f+1;
      printf("%lld\n",solve());
    }
    return 0;
}