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// https://vjudge.csgrandeur.cn/problem/HDU-1576
# include <bits/stdc++.h>
using namespace std ;
typedef long long LL ;
/*
测试用例
2
1000 53
87 123456789
答案
7922
6060
总结：
1、如果使用费马小定理+快速幂求逆元， ，
*/
const int MOD = 9973 ;
LL qmi ( LL a , LL k , LL p ) {
LL res = 1 ;
while ( k ) {
if ( k & 1 ) res = res * a % p ; //如果res是int,可能会越界
k > > = 1 ;
a = a * a % p ; //如果a是int, 可能会越界
}
return res ;
}
int main ( ) {
int T ;
cin > > T ;
while ( T - - ) {
LL a , b ;
cin > > a > > b ;
LL x ;
x = qmi ( b , MOD - 2 , MOD ) ; // b理解为分母,同余MOD的逆元,变除法为乘法
LL res = a * x % MOD ; // a* b在模MOD下的逆元 x
printf ( " %lld \n " , res ) ;
}
return 0 ;
}
