#include <bits/stdc++.h>

using namespace std;

ifstream in("ssnd.in");
ofstream out("ssnd.out");

#define pii pair<int, int>
#define pb push_back
#define fi first
#define se second

const int NMAX = 1000030;
const int INF = 0x3f3f3f3f;

int t;
long long n;
bool ciur[NMAX];
vector<int> prime;

void read()
{
    in >> t;
}

void eratostene()
{
    ciur[0] = ciur[1] = 1;

    for (int i = 4; i <= NMAX; i += 2)
        ciur[i] = 1;

    for (int i = 3; i * i <= NMAX; i += 2)
        if (!ciur[i]) // prim
            for (int j = 2; j * i <= NMAX; j++)
                ciur[i * j] = 1;

    for (int i = 2; i <= NMAX; i++)
        if (!ciur[i])
            prime.pb(i);
}

void solve()
{
    eratostene();
    while (t--)
    {
        in >> n;
        int i = 0, card = 1, divsum = 1;
        while (n > 1)
        {
            int p = prime[i], pow = 0;
            while (n % p == 0)
            {
                pow++;
                n /= p;
            }
            if (pow) // exista divizorul prim
            {
                // card = (d1+1)(d2+1)...(dn+1) //unde di este puterea divizorului prim pi din descompunerea in fac. primi
                card = card * (pow + 1);
                // sum = produs((pi^(di+1) - 1) / (pi - 1 ) )
                // ar fi nevoie de invers modular, sau folosesc formula pt a^n-b^n, asadar
                // sum = produs((p^di+p^(di-1)+..+p^2+p^1+1)), pt fiecare divizor
                int powp = p, currsum = 1;
                while (pow)
                    currsum = (currsum + powp) % 9973, powp = (powp * p) % 9973, pow--;

                divsum = (divsum * currsum) % 9973;
            }
            i++;
        }
        out << card << ' ' << divsum << '\n';
    }
}

int main()
{
    read();
    solve();

    return 0;
}