1 /**
2 大意：给定一组ai,bi . m = a1^b1 *a2^b2 * a3^ b3 * a4^b4*...*ai^bi
3 求最小的x!%m =0
4 思路: 将ai 质因子分解，若是x!%m=0 那么x！ 质因子分解之后 质因子的个数一定大于等于m的个数。二分求解可得
5 注意： 二分时，需要将，上下限 设定好，low =0； high = 1ll<<60;
6 **/
7
8 #include <iostream>
9 #include <cstring>
10 #include <cmath>
11 using namespace std;
12
13 long long pri[150];
14
15 void getApri(long long a,long long b){
16 int cnt;
17 for(int i=2;i*i<=a;i++)if(a%i==0){
18 cnt =0;
19 while(a%i==0){
20 cnt++;
21 a =a/i;
22 }
23 pri[i] += cnt*b;
24 if(a==1)
25 break;
26 }
27 if(a>1)
28 pri[a] += b;
29 }
30
31 long long cal(long long i,long long x){ x!中有多少个i因子
32 long long ret =0;
33 while(x){
34 x = x/i;
35 ret += x;
36 }
37 return ret;
38 }
39
40 bool judge(long long x){
41 for(int i=1;i<=100;i++)if(pri[i]){
42 long long tmp = cal(i,x);
43 if(tmp<pri[i])
44 return false;
45 }
46 return true;
47 }
48
49 int main(){
50 int t;
51 cin>>t;
52 while(t--){
53 memset(pri,0,sizeof(pri));
54 int n;
55 cin>>n;
56 long long a,b;
57 while(n--){
58 cin>>a>>b;
59 getApri(a,b);
60 }
61 long long low = 0,high = 1ll<<60;
62 long long ans;
63 while(low<=high){
64 long long mid = (low+high)/2;
65 if(judge(mid)){
66 ans = mid;
67 high = mid-1;
68 }
69 else
70 low = mid +1;
71 }
72 cout<<ans<<endl;
73 }
74
75 return 0;
76 }
