1 //2016.8.9
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<map>
 5 #include<algorithm>
 6 #include<cmath>
 7
 8 using namespace std;
 9
10 typedef long long ll;
11 const int N = 100005;
12 int dp[N][20];//d[i][j]表示从第i个数字开始向后2^j个数字这段区间内的gcd,具有递减性
13 map<int, ll> mp;
14
15 void init_rmq(int n)//初始化dp,求出每段区间的gcd
16 {
17     for(int j = 1; j < (int)log2(n)+1; j++)
18       for(int i = 1; i <= n; i++)
19       {
20           if(i+(1<<j)-1 <= n)
21             dp[i][j] = __gcd(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
22       }
23 }
24
25 int rmq(int l, int r)//查询
26 {
27     int k = (int)log2(r-l+1);
28     return __gcd(dp[l][k], dp[r-(1<<k)+1][k]);
29 }
30
31 int main()
32 {
33     int n, q, l, r, T, kase = 0;
34     cin>>T;
35     while(T--)
36     {
37         printf("Case #%d:\n", ++kase);
38         cin>>n;
39         mp.clear();
40         for(int i = 1; i <= n; i++)
41           scanf("%d", &dp[i][0]);
42         init_rmq(n);
43
44 //利用二分求具有相同gcd区间的数目
45 //-----------------------------------------------------------------------------------
46         for(int i = 1; i <= n; i++)
47         {
48             int a = i, b = n, mid, tmp, vs;
49             while(1)
50             {
51                 tmp = a;
52                 vs = rmq(i, a);
53                 while(a <= b)
54                 {
55                     mid = (a+b)>>1;
56                     if(rmq(i, mid)<vs) b = mid-1;
57                     else a = mid+1;
58                 }
59                 mp[vs]+=1ll*(b-tmp+1);
60                 b = n;
61                 if(a>b)break;
62             }
63         }
64 //------------------------------------------------------------------------------------
65
66         cin>>q;
67         while(q--)
68         {
69             scanf("%d%d", &l, &r);
70             int ans = rmq(l, r);
71             cout<<ans<<" "<<mp[ans]<<endl;
72         }
73     }
74
75     return 0;
76 }