#include<fstream>
#include<bitset>
using namespace std;

const int MOD = 9973;
int poz = 1;
long long primes[1000005];
bitset <1000005> v;

long long int suma, nrdiv;

int fastpower(int a,int p)
{
    int sol = 1;
    a %= MOD; /// %mod is required since numbers get can too large to fit into long long
    while(p)
    {
        if(p & 1)   sol = (sol * a) % MOD; /// by p&1 we verify parity in faster way
        a = (a * a) % MOD;
        p >>= 1; /// multiply by 2
    }
    return sol;
}


/// puts all the primes into Prim array, until 1 million there are around 78000 primes
void Sieve()
{
    v[0] = 1;
    v[1] = 1;
    primes[poz++] = 2;
    for (int i = 4;i <= 1000000;i += 2)
        v[i] = 1;
    for (int i = 3;i <= 1000000;i += 2)
    {
        if (v[i] == 0)
        {
            primes[poz++] = i;
            for (int j = 2 * i;j <= 1000000;j += i)
                v[j] = 1;
        }
    }
}

int invers_mod(int a)
{
    return fastpower(a, MOD - 2);
}

void suma_numar_div(long long int n)
{
    long long int cnt = 0;
    suma = 1;
    nrdiv = 1;
    /// we multiply primes[i]* primes[i] by 1LL since primes[i]*primes[i] can be over int
    for(int i = 1; i < poz && 1LL * primes[i]* primes[i] <= n ; i++) /// poz means number of primes
    {
        if(n % primes[i] == 0)
        {
            cnt = 0;
            while(n % primes[i] == 0)
            {
                cnt++;
                n /= primes[i];
            }
            nrdiv *= 1LL * (cnt + 1) % MOD;
            nrdiv %= MOD;
            suma *= (fastpower(primes[i],cnt + 1) - 1) / (primes[i] - 1);
        }
    }
    if(n > 1) /// that means we have 1 more prime factor
    {
        nrdiv = 2LL * nrdiv % MOD;
        suma = 1LL * suma * (n + 1) % MOD;
    }
}

int main()
{
    ifstream fin("ssnd.in");
    ofstream fout("ssnd.out");

    long long t,n;
    /// This program calculates sum of the divisors of T (T<1000) numbers and each number can be max 1 billion!!!
    Sieve();
    fin >> t;
    while(t--)
    {
        fin >> n;
        suma_numar_div(n);
        fout << nrdiv << " " << suma << '\n';
    }
    return 0;
}