#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<stack>
#include<bitset>
#include<cstdlib>
#include<cmath>
#include<set>
#include<list>
#include<deque>
#include<map>
#include<queue>
#define ll long long int
using namespace std;
inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
inline ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
int moth[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
int dir[4][2]={1,0 ,0,1 ,-1,0 ,0,-1};
int dirs[8][2]={1,0 ,0,1 ,-1,0 ,0,-1, -1,-1 ,-1,1 ,1,-1 ,1,1};
const int inf=0x3f3f3f3f;
const ll mod=1e9+7;
ll mx[11][1007],mn[11][1007];
int a[10]={6,2,5,5,4,5,6,3,7,6}; //每个字符消耗的火柴数
ll dp[100][1000]; //前i个数字 用了j根火柴
int b[100]; //位数
void init(){for(int i=0;i<11;i++)for(int j=0;j<1007;j++){mx[i][j]=-1; //i位数 用j个火柴可以达到的最大值 mn[i][j]=1e17;//i位数 用j个火柴可以达到的最小值
        }mx[0][0]=mn[0][0]=0;for(int i=1;i<11;i++)for(int j=0;j<=i*7;j++)for(int k=0;k<=9;k++){if(j<a[k]) continue;mx[i][j]=max(mx[i][j],mx[i-1][j-a[k]]*10+k); //类似背包找最大值 mn[i][j]=min(mn[i][j],mn[i-1][j-a[k]]*10+k);}
}
int main(){ios::sync_with_stdio(false);init();int t;cin>>t;while(t--){memset(dp,-1,sizeof(dp));memset(b,0,sizeof(b));int n;cin>>n;string s;cin>>s;int len=s.length();int m=0; //火柴数int k=1; //字符数for(int i=0;i<len;i++){if(s[i]=='+'){k++;m+=2;}else if(s[i]=='-'){k++;m++;}else{m+=a[s[i]-'0'];b[k]++;    }}for(int i=0;i<=m;i++)if(mx[b[1]][i]!=-1)dp[1][i]=mx[b[1]][i]; //初始化边界 for(int i=2;i<=k;i++)for(int j=0;j<=m;j++)for(int p=1;p<=700;p++){if(j-p-2>=0&&dp[i-1][j-p-2]!=-1){ //考虑+号 if(mx[b[i]][p]!=-1)dp[i][j]=max(dp[i][j],dp[i-1][j-p-2]+mx[b[i]][p]);}if(j-p-1>=0&&dp[i-1][j-p-1]!=-1){ //考虑-号 if(mn[b[i]][p]!=1e17)dp[i][j]=max(dp[i][j],dp[i-1][j-p-1]-mn[b[i]][p]);}}cout<<dp[k][m]<<endl;}return 0;
}