1 #include <cstdio>
  2 #include <iostream>
  3 #include <algorithm>
  4 #include <vector>
  5 #include <set>
  6 #include <map>
  7 #include <string>
  8 #include <cstring>
  9 #include <stack>
 10 #include <queue>
 11 #include <cmath>
 12 #include <ctime>
 13 #include <bitset>
 14 #include <utility>
 15 #include <assert.h>
 16 using namespace std;
 17 #define rank rankk
 18 #define mp make_pair
 19 #define pb push_back
 20 #define xo(a,b) ((b)&1?(a):0)
 21 #define tm tmp
 22 //#define LL ll
 23 typedef unsigned long long ull;
 24 typedef pair<int,int> pii;
 25 typedef long long ll;
 26 typedef pair<ll,int> pli;
 27 typedef pair<ll,ll> pll;
 28 const int INF=0x3f3f3f3f;
 29 const ll INFF=0x3f3f3f3f3f3f3f3fll;
 30 const int MAX=1e5+5;
 31 //const ll MAXN=2e8;
 32 const int MAX_N=MAX;
 33 const ll MOD=998244353;
 34 //const long double pi=acos(-1.0);
 35 //const double eps=0.00000001;
 36 int gcd(int a,int b){return b?gcd(b,a%b):a;}
 37 template<typename T>inline T abs(T a) {return a>0?a:-a;}
 38 template<class T> inline
 39 void read(T& num) {
 40     bool start=false,neg=false;
 41     char c;
 42     num=0;
 43     while((c=getchar())!=EOF) {
 44         if(c==‘-‘) start=neg=true;
 45         else if(c>=‘0‘ && c<=‘9‘) {
 46             start=true;
 47             num=num*10+c-‘0‘;
 48         } else if(start) break;
 49     }
 50     if(neg) num=-num;
 51 }
 52 inline ll powMM(ll a,ll b,ll M){
 53     ll ret=1;
 54     a%=M;
 55 //    b%=M;
 56     while (b){
 57         if (b&1) ret=ret*a%M;
 58         b>>=1;
 59         a=a*a%M;
 60     }
 61     return ret;
 62 }
 63 void open()
 64 {
 65 //    freopen("1009.in","r",stdin);
 66     freopen("out.txt","w",stdout);
 67 }
 68 const int MAXN=100005;
 69 int t1[MAXN],t2[MAXN],c[MAXN];//求SA数组需要的中间变量，不需要赋值
 70 //待排序的字符串放在s数组中，从s[0]到s[n-1],长度为n,且最大值小于m,
 71 //除s[n-1]外的所有s[i]都大于0，r[n-1]=0
 72 //函数结束以后结果放在sa数组中
 73 bool cmp(int *r,int a,int b,int l)
 74 {
 75     return r[a] == r[b] && r[a+l] == r[b+l];
 76 }
 77 void da(int str[],int sa[],int rank[],int height[],int n,int m)
 78 {
 79     str[n++]=0;
 80     int i, j, p, *x = t1, *y = t2;
 81     //第一轮基数排序，如果s的最大值很大，可改为快速排序
 82     for(i = 0;i < m;i++)c[i] = 0;
 83     for(i = 0;i < n;i++)c[x[i] = str[i]]++;
 84     for(i = 1;i < m;i++)c[i] += c[i-1];
 85     for(i = n-1;i >= 0;i--)sa[--c[x[i]]] = i;
 86     for(j = 1;j <= n; j <<= 1)
 87     {
 88         p = 0;
 89         //直接利用sa数组排序第二关键字
 90         for(i = n-j; i < n; i++)y[p++] = i;//后面的j个数第二关键字为空的最小
 91         for(i = 0; i < n; i++)if(sa[i] >= j)y[p++] = sa[i] - j;
 92         //这样数组y保存的就是按照第二关键字排序的结果
 93         //基数排序第一关键字
 94         for(i = 0; i < m; i++)c[i] = 0;
 95         for(i = 0; i < n; i++)c[x[y[i]]]++;
 96         for(i = 1; i < m;i++)c[i] += c[i-1];
 97         for(i = n-1; i >= 0;i--)sa[--c[x[y[i]]]] = y[i];
 98         swap(x,y);
 99         //根据sa和x数组计算新的x数组 swap(x,y);
100         p = 1; x[sa[0]] = 0;
101         for(i = 1;i < n;i++)
102             x[sa[i]] = cmp(y,sa[i-1],sa[i],j)?p-1:p++;
103         if(p >= n)break;
104         m = p;//下次基数排序的最大值
105     }
106     int k = 0; n--;
107     for(i = 0;i <= n;i++)rank[sa[i]] = i;
108     for(i = 0;i < n;i++)
109     {
110         if(k)k--;
111         j = sa[rank[i]-1]; while(str[i+k] == str[j+k])k++; height[rank[i]] = k;
112     }
113 }
114 int rank[MAXN],height[MAXN];
115 int RMQ[MAXN];
116 int mm[MAXN];
117 int best[20][MAXN];
118 void initRMQ(int n)
119 {
120     mm[0]=-1;
121     for(int i=1;i<=n;i++)
122     mm[i]=((i&(i-1))==0)?mm[i-1]+1:mm[i-1];
123     for(int i=1;i<=n;i++)best[0][i]=i;
124     for(int i=1;i<=mm[n];i++)
125     for(int j=1;j+(1<<i)-1<=n;j++)
126     {
127         int a=best[i-1][j];
128         int b=best[i-1][j+(1<<(i-1))];
129         if(RMQ[a]<RMQ[b])best[i][j]=a; else best[i][j]=b;
130     }
131 }
132 int askRMQ(int a,int b)
133 {
134     int t; t=mm[b-a+1];
135     b-=(1<<t)-1;
136     a=best[t][a];b=best[t][b];
137     return RMQ[a]<RMQ[b]?a:b;
138 }
139 int lcp(int a,int b)
140 {
141     a=rank[a];b=rank[b];
142     if(a>b)swap(a,b);
143     return height[askRMQ(a+1,b)];
144 }
145 string sts[MAXN];
146 int st[MAXN],sa[MAXN];
147 
148 struct Trie {
149     bool isWord;
150     Trie* child[27];
151     Trie(bool isWord):isWord(isWord)
152     {
153         memset(child,0,sizeof(child));
154     }
155     void addWord(string &s)
156     {
157         Trie*cur =this;
158         for(char c: s)
159         {
160             Trie* next=cur->child[c-‘a‘+1];
161             if(next==nullptr)
162                 next=cur->child[c-‘a‘+1]=new Trie(false);
163             cur=next;
164         }
165         cur->isWord=true;
166     }
167     int checkstr(string s)
168     {
169         Trie*cur=this;int re=0;
170         for(char c:s)
171         {
172             Trie* next=cur->child[c-‘a‘+1];
173             if(next==nullptr)break;
174             else{cur=next;++re;}
175         }
176         return re;
177     }
178     ~Trie()
179     {
180         for(int i=0;i<27;++i)
181         {
182             if(child[i])
183                 delete child[i];
184         }
185     }
186 };
187 int t,n,m,x,y,an;
188 int main()
189 {
190     scanf("%d",&t);
191     while(t--)
192     {
193         scanf("%d",&n);Trie now(false);
194         for(int i=1;i<=n;i++){cin>>sts[i];now.addWord(sts[i]);}
195         scanf("%d",&m);
196         for(int i=1;i<=m;i++)
197         {
198             scanf("%d%d",&x,&y);int len1=sts[x].length(),len2=sts[y].length();
199             for(int i=0;i<len1;i++)st[i]=sts[x][i]-‘a‘+1;st[len1]=27;
200             for(int i=0;i<len2;i++)st[len1+1+i]=sts[y][i]-‘a‘+1;
201             da(st,sa,rank,height,len1+len2+1,28);an=0;
202             for(int i=1;i<=len1+len2+1;i++)
203             {
204                 int lo1=sa[i],lo2=sa[i-1];if(lo1>lo2)swap(lo1,lo2);
205                 if(height[i]>an&&lo1<len1&&lo2>len1)
206                 {
207                     int tem=now.checkstr(sts[x].substr(lo1,height[i]));
208                     if(tem>an)an=tem;
209                 }
210             }
211             printf("%d\n",an);
212         }
213     }
214 }