LL cnt = 0;

        for(int i=0; i < ans && primes[i] * primes[i] <= S; i++)    //ans是primes中质数的个数

        {

            LL cnt2 = 0;                                 

            while(S % primes[i] == 0)

            {

                cnt2++;

                S /= primes[i];

 

            }

            if(S > 0){

                base[cnt] = primes[i]; //base中是底数

                mi[cnt++] = cnt2; //cnt2是幂的个数

            }

        }