#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
#include <set>
using namespace std;

#define MM(a) memset(a,0,sizeof(a))

typedef long long LL;
typedef unsigned long long ULL;
const int maxn = 1e4+5;
const int mod = 1000000007;
const double eps = 1e-7;
bool prime[maxn];
int p[maxn],k;
///素数筛选
void isprime()
{
    k = 0;
    MM(prime);
    for(int i=2; i<maxn; i++)
    {
        if(!prime[i])
        {
            p[k++] = i;
            for(int j=i*i; j<maxn; j+=i)
                prime[j] = 1;
        }
    }
}
int fac[maxn], num[maxn], cnt;
///分解素因子算法
void Dec(int x)
{
    MM(num);
    cnt = 0;
    for(int i=0; p[i]*p[i]<=x&&i<k; i++)
    {
        if(x%p[i] == 0)
        {
            fac[cnt] = p[i];
            while(x%p[i] == 0)
            {
                num[cnt]++;
                x /= p[i];
            }
            cnt++;
        }
    }
    if(x > 1)
    {
        fac[cnt] = x;
        num[cnt++] = 1;
    }
}
///欧拉公式的延伸：一个数的所有质因子之和是euler(n)*n/2
int Euler(int m)
{
    int ret = m;
    for(int i=2; i*i<=m; i++)
    {
        if(m%i == 0)
            ret -= ret/i;
        while(m%i == 0)
            m /= i;
    }
    if(m > 1)
        ret -= ret/m;
    return ret;
}
///最大公约数
LL gcd(LL x, LL y)
{
    if(y == 0)
        return x;
    return gcd(y, x%y);
}
///扩展欧几里得算法
void exgcd(LL a, LL b, LL &x, LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return;
    }
    exgcd(b, a%b, x, y);
    LL tmp = x;
    x = y;
    y = tmp-(a/b)*y;
}

int main()
{
    int m, t;
    //isprime();
    cin>>t;
    while(t--)
    {
///素因子分解
        /**
        cin>>m
        Dec(m);
        for(int i=0; i<cnt-1; i++)
            for(int j=0; j<num[i]; j++)
                cout<<fac[i]<<'*';
        for(int i=0; i<num[cnt-1]-1; i++)
            cout<<fac[cnt-1]<<"*";
        cout<<fac[cnt-1]<<endl;
        **/
///欧拉函数
        /**cout<<Euler(m)<<endl;**/

///扩展欧几里得算法
        /**
        LL a, b, x, y;
        cin>>a>>b;
        if(gcd(a,b) != 1)///没有解
        {
            puts("Sorry");
            continue;
        }
        exgcd(a, b, x, y);
        x = (b + x % b) % b;
        y = (1 - a * x) / b;
        cout<< x << " " << y << endl;
        **/
    }
    return 0;
}