#include <bits/stdc++.h>

 #define MAX 10000

 #define INF 100000000

 #define LL long long

 #define U unsigned

 using namespace std;

 int cas=,T;

 U int prime[],pn,*r,rn,*rr,n,*rrp;

 struct node

 {

     U int x,y;

     bool operator<(const node&a)const

     {

         return x>a.x;

     }

 };

 /*void init()        //TLE

 {

     pn=1;

     bool *vis=new bool[INF+10];

     priority_queue<node>q;

     for(int i=2;i<=MAX;i++)

         if(!vis[i])

         {

             for(int j=i*i;j<=INF;j+=i) vis[j]=1;

             q.push(node{i,i});

         }

     prime[0]=1;

     for(U int i=MAX+1,x=1;i<=INF;i++)

     {

         if(!vis[i]) prime[pn++]=i;

     }

     prime[pn++]=INF+10;

     delete []vis;

     r=new U int[pn+10];

     r[0]=1;

     for(int i=1;i<pn;i++) r[i]=r[i-1]*prime[i];

     U int res=1;

     rr=new U int[3000];

     rrp=new U int[3000];

     rr[0]=rrp[0]=1;

     rn=1;

     while(!q.empty())

     {

         node u=q.top();q.pop();

         rrp[rn]=u.x;

         rr[rn++]=res*u.y;

         if((U LL)u.x*u.y<=INF) q.push(node{u.x*u.y,u.y});

     }

     rrp[rn++]=INF+10;

     for(int i=1;i<rn;i++) rr[i]=rr[i-1]*rr[i];

 }

 */

 void init()            //最大数为10000^2，则大于10000的素数最小公倍数的最大次幂为1，小于一万时不定

 {

     bool *vis=new bool[INF+];

     pn=;

     for(U int i=;i<=INF;i++)        //O(n)筛法

     {

         if(!vis[i]) prime[pn++]=i;

         for(U int j=;j<pn&&i*prime[j]<=INF;j++)

         {

             vis[i*prime[j]]=;

             if(i%prime[j]==) break;

         }

     }

     delete []vis;            //内存不足，用完即删

     priority_queue<node>q;

     r=new U int[pn];

     for(int i=;i<pn;i++)            //prime存质数，r存第i个质数对应的结果（还没算1~10000的部分）

     {

         if(prime[i]<=MAX) { r[i]=;q.push(node{prime[i],prime[i]}); }

         else r[i]=r[i-]*prime[i];

     }

     U int res=;

     rr=new U int[];

     rrp=new U int[];

     rr[]=rrp[]=;

     rn=;

     while(!q.empty())            //rr存质数的幂次升序，rrp存底数

     {

         node u=q.top();q.pop();

         rrp[rn]=u.x;

         rr[rn++]=res*u.y;

         if((U LL)u.x*u.y<=INF) q.push(node{u.x*u.y,u.y});

     }

     rrp[rn++]=INF+;

     for(int i=;i<rn;i++) rr[i]=rr[i-]*rr[i];        //将rrp乘起来作为1~10000的结果

 }

 int main()

 {

     //printf("%d\n",sizeof(vis)+sizeof(prime));

     init();

 //    for(int i=0;i<rn;i++) printf("%u %u\n",rrp[i],rr[i]);

     //freopen("1.in","w",stdout);

     //freopen("1.in","r",stdin);

     //freopen("1.out","w",stdout);

     //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);

     scanf("%d",&T);

     while(T--)

     {

         scanf("%u",&n);

         int i1=upper_bound(prime,prime+pn,n)--prime;        //查找>10000的结果

         int i2=upper_bound(rrp,rrp+rn,n)--rrp;                //查找1~10000的结果

         printf("%u\n",r[i1] * rr[i2]);

     }

     delete []r;

     delete []rrp;

     delete []rr;

     return ;

 }