#include <fstream>
#include <math.h>
using namespace std;
int m,numar,n,set,x,t,i,j,a[100000];
unsigned long long prod_sus,prod_jos,prod;
bool prim[1000000];
long long nr;
int putere(int num,int p)
{
    int i,sol=1;
    for(i=0;(1<<i)<=p;++i)
    {
        if (((1<<i)&p)>0) sol=(sol*num);
        num=(num*num);
    }
    return sol;
}
int main()
{
    int p;
    ifstream fi("ssnd.in");
    ofstream fo("ssnd.out");
    m=9973;
    n=0;
    for(i=2;i<=1000000;i++)
       if(!prim[i]) { a[++n]=i; for(j=2*i;j<=1000000;j+=i) prim[j]=1; }
    fi>>t;
    for(set=1;set<=t;set++)
    {
       fi>>nr;
       //x=(int)sqrt(nr);
       prod_sus=1; numar=1;
       prod_jos=1;
       i=1;
       for(i=1;i<=n&&a[i]<=nr;i++)
       if(nr%a[i]==0)
       {
           p=0;
           while(nr%a[i]==0) { nr/=a[i]; p++; }

           prod_sus=prod_sus*(putere(a[i],p+1)-1);
           prod_jos=prod_jos*(a[i]-1);
           numar*=(p+1);
       }

       prod=(prod_sus/prod_jos)%m;
       if(numar==prod==1) fo<<"2 "<<nr+1<<"\n"; else
       fo<<numar<<" "<<(int)prod<<"\n";
    }
    fo.close();
    return 0;
} /*
#include <fstream>

using namespace std;

const int MAX_N = 1000005;
const int MOD = 9973;

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

long long N;
int T, P[MAX_N], K;
char viz[MAX_N];

void ciur()
{
	for(int i = 2; i < MAX_N; ++i)
		if(viz[i] == 0)
		{
			P[++K] = i;

			for(int j = i+i; j < MAX_N; j += i)
				viz[j] = 1;
		}
}

inline int pow(int x, int p)
{
	int rez = 1;
	x %= MOD;

	for(; p; p >>= 1)
	{
		if(p & 1)
			rez *= x,
			rez %= MOD;
		x *= x;
		x %= MOD;
	}

	return rez;
}
void solve()
{
	fin >> N;

	int	nd = 1, sd = 1;

	long long n = N;
	for(int i = 1; 1LL * P[i] * P[i] <= N; ++i)
	{
		if(N % P[i]) continue;
		int p = 0, s = 1;
		while(n % P[i] == 0)
			n /= P[i], s *= P[i], ++p;

		nd *= (p+1);
		sd *= ((1LL*s*P[i] - 1) % MOD);
		sd %= MOD;
		sd *= pow(P[i]-1, MOD-2);
		sd %= MOD;
	}

	if(n > 1)
	{
		nd *= 2;
		sd *= ((1LL*n*n - 1) % MOD);
		sd %= MOD;
		sd *= pow(n-1, MOD-2);
		sd %= MOD;
	}

	fout << nd << " " << sd << "\n";
}

int main()
{
	ciur();

	for(fin >> T; T; --T)
		solve();
}
*/
/*
6 28
36 6458
12 684
12 1726
8 6989
4 168
2 9252
6 78
2 68
6 28

*/