#include <iostream>
#include <fstream>
#define mod 9973
#define maxn 1000010
#include <cstring>
using namespace std;
ifstream in("ssnd.in");
ofstream out("ssnd.out");
bool c[maxn];
int prime[maxn];
int ct=1;
void ciur()
{
    int i,j;
    //c[0]=1;
    //c[1]=1;
    prime[1]=2;
    for(i=3;i*i<maxn;i+=2)
        if(c[i]==0)
            {
                prime[++ct]=i;
                for(j=i*i;j<maxn;j=j+2*i)
                    c[j]=1;
            }

      for(;i<maxn;i+=2)
        if(c[i]==0)
            prime[++ct]=i;
}

long long lgpow(long long b,long long p)
{
    if(p==0)
        return 1;
    if(p==1)
        return b%mod;
    if(p%2==0)
        return (lgpow((long long)(b*b)%mod,p/2))%mod;
    else
        return (b*lgpow((long long)(b*b)%mod,(p-1)/2))%mod;
}

void solve(long long x)
{
    long long i;
    long long d;
    long long nr=1,sum=1;

        for(i=1;i<=ct&&1LL*prime[i]*prime[i]<=x;i++)
        {

                if(x%prime[i]==0)
                {
                    d=0;
                    while(x%prime[i]==0)
                    {
                        x=x/prime[i];
                        d++;
                    }
                    nr=nr*(d+1);
                    int pow1=( lgpow( prime[i] , d+1 ) - 1 + mod) % mod;
                    int pow2=( lgpow( prime[i]-1 , mod-2 ) ) % mod;
                    sum=(((sum*pow1)%mod)*pow2)%mod;
                }
            }
        if(x>1)
        {
            nr*=2;
            sum=(1LL*sum*(x+1))%mod;
        }

        out<<nr<<" "<<sum%mod<<"\n";



/*

        sd=1;nd=1;
        scanf("%lld",&n);
        for(i=1;i<=np&&P[i]*P[i]<=n;i++)
            if(n%P[i]==0)
            {
                sum=1;cnt=1;term=1;
                prim=P[i]%9973;
                while(n%P[i]==0)
                {
                    cnt++;
                    term=(term*prim)%9973;
                    sum=(sum+term)%9973;
                    n/=P[i];
                }
                sd=(sd*sum)%9973;
                nd*=cnt;
            }

*/



}


int main()
{
    int t;
    long long n;
    in>>t;
    ciur();
    while(t--)
    {
        in>>n;
        solve(n);
    }


    return 0;
}
//4 8304