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#include <fstream>
#include <math.h>
//#include <iostream>
std::ifstream f("ssnd.in");
std::ofstream g("ssnd.out");
using namespace std;
constexpr int MAXN = 1000001;
//(a - b) mod p = ((a mod p - b mod p) + p) mod p
//(a / b) mod p = ((a mod p) * (b^(-1) mod p)) mod p
//Where b^(-1) mod p is the modular inverse of b mod p. For p = prime, b^(-1) mod p = b^(p - 2) mod p
void findPrimes(long long primes[], long int &K)
{
bool filter[MAXN];
for(auto i=0; i<MAXN; ++i)
{
filter[i] = true;
}
K = 0;
for(auto i=2; i<=MAXN; ++i)
{
if(filter[i])
{
primes[K] = i;
++K;
for(auto j=i+i; j<MAXN; j+=i)
{
filter[j] = false;
}
}
}
}
long long power(long long n, long long p, long long mod)
{
auto r = 1LL;
while( p != 1 )
{
if( p % 2 ==1 )
r = (n*r) % mod;
n = (n*n) % mod;
p = p / 2;
}
return ( n * r ) % mod;
}
void computeValues(long long &nr, long long &sum, long long nr_d, long long prime)
{
auto mi = power(prime-1, 9971, 9973); ///For p = prime, b^(-1) mod p = b^(p - 2) mod p
//std::cout<< mi <<" " << mi2 << '\n';
auto prime_pow = power(prime, nr_d + 1, 9973) - 1;
//std::cout<<prime_pow<<" " <<prime_pow1 << '\n';
sum = (sum * prime_pow * mi) % 9973;
nr *= nr_d + 1;
}
long long primes[100000];
int main()
{
int t=0;
long long x=0;
long long int sum = 0;
long long int nr = 0;
long int K;
findPrimes(primes, K);
f>>t;
while(t > 0)
{
f >> x;
sum = 1;
nr = 1;
for(auto i = 0L; i < K; ++i)
{
auto nr_d = 0;
while( x % primes[i] == 0 )
{
++ nr_d;
x /= primes[i];
}
if(nr_d > 0)
{
computeValues(nr, sum, nr_d, primes[i]);
}
}
if(x > 1)
{
++nr;
computeValues(nr, sum, 1, x);
}
g << nr << ' ' << sum <<'\n';
--t;
}
}