1 /*
 2 Author:    Leo.W
 3 Descriptipn:    計(jì)算N^N結(jié)果的最左邊的數(shù)            
 4 How to Do:    數(shù)學(xué)題 由sum=N^N,兩邊對(duì)10取對(duì)數(shù),log10(sum)=Nlog10(N),有sum=10^(Nlog10(N));
 5             由于10的整數(shù)次冪首位均為1，則僅需考慮Nlog10(N)的結(jié)果的小數(shù)部分即可
 6   */
 7 #include <stdio.h>
 8 #include <math.h>
 9 int main(){
10     //freopen("in.txt","r",stdin);
11     int t;
12     __int64 num,sum2;
13     scanf("%d",&t);
14     while(t--){
15         scanf("%I64d",&num);
16         double sum1=num*log10(double(num));
17         sum2=(__int64)sum1;
18         sum1-=sum2;
19         num=(__int64)pow(10.0,sum1);
20         printf("%I64d\n",num);
21     }
22     return 0; 
23 }