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.

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int kMaxN = 1005;
const int kMaxM = 6;

long long f[kMaxN][kMaxM];
long long g[kMaxN][kMaxM];
int a[kMaxN];
char curse[kMaxM];

int main() {
  int T;
  scanf("%d", &T);
  for (int cas = 1; cas <= T; ++cas) {
    int n, m, k;
    scanf("%d %d %d", &n, &m, &k);
    memset(f, 0xa0, sizeof(f));
    memset(g, 0x70, sizeof(g));
    for (int i = 0; i <= n; ++i) {
      f[i][0] = k;
      g[i][0] = k;
    }
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
    }
    scanf("%s", curse);
    for (int j = 1; j <= m; ++j) {
      for (int i = j; i <= n; ++i) {
        f[i][j] = f[i - 1][j];
        g[i][j] = g[i - 1][j];
        if (curse[j - 1] == ''+'') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] + a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] + a[i]);
        } else if (curse[j - 1] == ''-'') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] - a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] - a[i]);
        } else if (curse[j - 1] == ''*'') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] * a[i]);
          f[i][j] = max(f[i][j], g[i - 1][j - 1] * a[i]);
          g[i][j] = min(g[i][j], f[i - 1][j - 1] * a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] * a[i]);
        } else if (curse[j - 1] == ''/'') {
          f[i][j] = max(f[i][j], f[i - 1][j - 1] / a[i]);
          f[i][j] = max(f[i][j], g[i - 1][j - 1] / a[i]);
          g[i][j] = min(g[i][j], f[i - 1][j - 1] / a[i]);
          g[i][j] = min(g[i][j], g[i - 1][j - 1] / a[i]);
        }
        //printf("[%d][%d] = %lld %lld\n", i, j, f[i][j], g[i][j]);
      }
    }
    printf("%lld\n", f[n][m]);
  }
  return 0;
}

#include<bits/stdc++.h>
#define per(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long ll;
//#define int long long
const ll inf =2333333333333333LL;
const double eps=1e-8;
int read(){
    char ch=getchar();
    int res=0,f=0;
    while(ch<''0'' || ch>''9''){f=(ch==''-''?-1:1);ch=getchar();}
    while(ch>=''0''&&ch<=''9''){res=res*10+(ch-''0'');ch=getchar();}
    return res*f;
}
// ------------------------head
#define mod 1000000009
const int siz=10005;
ll dp[1005][6],k;
int T,n,m,a[1005];
char ch[6];
ll fun(int rhs,ll d,char op){
    if(op==''+'')return d+(ll)rhs;
    else if(op==''-'')return d-(ll)rhs;
    else if(op==''*'')return d*(ll)rhs;
    else if(op==''/'')return d/(ll)rhs; 
}

signed main()
{
    scanf("%d",&T);
    while(T--){
        scanf("%d%d%lld",&n,&m,&k);
        ll ans=-inf;
        per(i,1,n)scanf("%d",&a[i]);
        per(i,1,m)cin>>ch[i];
        ll _mi[1005][6];
        per(i,0,n)dp[i][0]=_mi[i][0]=k;
        per(i,1,m){per(j,0,n)_mi[j][i]=inf;per(j,0,n)dp[j][i]=-inf;}
        per(i,1,n){
            dp[i][1]=max(dp[i-1][1],fun(a[i],k,ch[1]));
            _mi[i][1]=min(_mi[i-1][1],fun(a[i],k,ch[1]));
        }
        per(i,2,m){//区分正数和负数
            per(j,i,n){
                if(a[j]<0&&(ch[i]==''*''||ch[i]==''/'')){
                    dp[j][i]=max(dp[j-1][i],fun(a[j],_mi[j-1][i-1],ch[i]));
                    _mi[j][i]=min(_mi[j-1][i],fun(a[j],dp[j-1][i-1],ch[i]));
                }
                else {
                    ll ma=dp[j-1][i-1];
                    dp[j][i]=max(dp[j-1][i],fun(a[j],dp[j-1][i-1],ch[i]));
                    _mi[j][i]=min(_mi[j-1][i],fun(a[j],_mi[j-1][i-1],ch[i]));
                }
            }
        }
        if(m==1){
            ll res=-inf;
            per(i,1,n)res=max(res,dp[i][1]);
            printf("%lld\n",res);
        }
        else printf("%lld\n",dp[n][m]);
    }
    return 0;
}