/*  思路是很好想的，设f[i][j]为A染色体前i个基因和B染色体前j个基因匹配的最大值  第一次测样例WA了一把，f又没有赋最小值，今天第二次了，幸亏样例测出来了，不然又要WA一次。 */#include<cstdio>#include<iostream>#include<cstring>#define N 110using namespace std;int a[N],b[N],f[N][N],n,m;int v[5][5]={{5,-1,-2,-1,-3},             {-1,5,-3,-2,-4},             {-2,-3,5,-2,-2},             {-1,-2,-2,5,-1},             {-3,-4,-2,-1,0}};void work(){    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        char c;cin>>c;        if(c==‘A‘)a[i]=0;        if(c==‘C‘)a[i]=1;        if(c==‘G‘)a[i]=2;        if(c==‘T‘)a[i]=3;    }    scanf("%d",&m);    for(int i=1;i<=m;i++)    {        char c;cin>>c;        if(c==‘A‘)b[i]=0;        if(c==‘C‘)b[i]=1;        if(c==‘G‘)b[i]=2;        if(c==‘T‘)b[i]=3;    }    f[0][0]=0;    for(int i=0;i<=n;i++)      for(int j=0;j<=m;j++)      {          if(i)f[i][j]=max(f[i][j],f[i-1][j]+v[a[i]][4]);          if(j)f[i][j]=max(f[i][j],f[i][j-1]+v[4][b[j]]);          if(i&&j)f[i][j]=max(f[i][j],f[i-1][j-1]+v[a[i]][b[j]]);      }    printf("%d\n",f[n][m]);}int main(){    freopen("jh.in","r",stdin);    int T;scanf("%d",&T);    while(T--)    {        memset(f,466,sizeof(f));        work();    }    return 0;}