/*
    题意： 给出一些相连的管，从第一个管开始注水，问到达管ID的Y的时间
            No two links have the same y coordinates.
            所以最多一个pipe到另一个pipe
            a->b,而当b升到跟a一样高时，两者一起升
            利用这点，可以用一个栈保存现场！

            当level升到某一个link的高度时，就传给另一个pipe k了，更新level = py[k]+ph[k]，这时管数目为1
            （当然可以继续传给k'，用栈记录即可）
            而再次回到这个高度时，pop栈，一起升了，total+=栈顶

            即：传播开+回来一起升  Like Stack        
*/
#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;

int px[21],py[21],ph[21],pNum;
int lxl[51],lxr[51],ly[51],lNum;
bool water[21];

int main()
{
    //freopen("in","r",stdin);
    int T,Y,ID;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&pNum);
        for(int i=1;i<=pNum;i++)
            scanf("%d%d%d",&px[i],&py[i],&ph[i]);
        scanf("%d",&lNum);
        for(int i=1;i<=lNum;i++)
        {
            scanf("%d%d%d",&lxl[i],&ly[i],&lxr[i]);
            lxr[i]+=lxl[i]--;
        }
        scanf("%d%d",&ID,&Y);
        if(Y<=py[ID]||Y>py[ID]+ph[ID])//注意这种情况！
        {
            puts("No Solution");
            continue;
        }
        bool ns=0,New;
        int time=0,level=py[1]+ph[1],total=1,overflow=py[1];
        memset(water,0,sizeof(water));
        water[1]=1;
        vector<pair<int,int> >S;
        while(true)
        {
            New = 0;
            for(int i=1;i<=lNum;i++)
            {
                if(ly[i]==level)
                {
                    for(int j=1;j<=pNum;j++)
                    {
                        if(water[j])
                        {
                            int k;
                            for(k=1;k<=pNum;k++)
                            {
                                if(lxl[i]==px[j]&&lxr[i]==px[k]||lxl[i]==px[k]&&lxr[i]==px[j])
                                {
                                    if(!water[k])//new
                                    {
                                        S.push_back(make_pair(total,j));
                                        level=py[k]+ph[k];
                                        if(overflow<py[k])overflow=py[k];
                                        total=1;
                                        water[k]=1;
                                        New = 1;
                                    }
                                    else 
                                    {
                                        if(S.size()&&(S[S.size()-1].second==j||S[S.size()-1].second==k))//pop and back
                                        {
                                            total += S[S.size()-1].first;
                                            //printf("%d %d %d\n",j,k,total);
                                            S.pop_back();
                                        }
                                    }
                                    break;//find out ,so break
                                }
                            }
                            if(k<=pNum)break;
                        }
                    }
                    break;//only one link = level  (No two links have the same y coordinates.)
                }
            }
            if(New)continue;
            if(level<=overflow||Y<=overflow)//
            {
                ns=1;
                break;
            }
            if(level<=Y&&water[ID])break;
            level--;
            time+=total;
        }
        if(ns)puts("No Solution");
        else printf("%d\n",time);
    }
    return 0;
}
/*
Test

2
0 0 3
2 3 1
1
1 3 1
1 3

7
10 4 6
7 0 13
12 0 13
3 2 8
0 7 6
1 0 6
5 5 7
13
2 1 5
2 3 1
1 9 2
1 12 4
4 8 1
11 10 1
8 13 4
8 2 4
4 4 3
8 5 2
4 6 1
6 7 1
6 11 1
7 12

*/