You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of n^k.

Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing two integers: n (2 ≤ n < 2³¹) and k (1 ≤ k ≤ 10⁷).

For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that n^k contains at least six digits.

5

123456 1

123456 2

2 31

2 32

29 8751919

Case 1: 123 456

Case 2: 152 936

Case 3: 214 648

Case 4: 429 296

Case 5: 665 669

题意：输出n^k的前三位和后三位。

思路：后三位直接快速幂取模，前三位化为科学计数法取对数推导：

            n^k=a.bc*10^m ( m为n^k的位数，即m=(int)lg(n^k)=(int)(k*lgn) );

            求对数:  k*lgn=lg(a.bc)+m

             即 a.bc=10^(k*lgn-m)=10^(k*lgn-(int)(k*lgn));

              abc=a.bc*100;

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<vector>
#include<stack>
#include<queue>
#include<set>
#include<map>
#include<string>
#include<math.h>
#include<cctype>

using namespace std;

typedef long long ll;
const int maxn=1000100;
const int INF=(1<<29);
const double EPS=0.0000000001;
const double Pi=acos(-1.0);
const int p=1000;

ll n,k;

ll qpow(ll n,ll k)
{
    ll res=1;
    while(k){
        if(k&1) res=(res%p)*(n%p)%p;
        n=(n%p)*(n%p)%p;
        k>>=1;
    }
    return res;
}

ll f(ll n)
{
    double x=k*log10(n)-(int)(k*log10(n));
    return pow(10,x)*100;
}

int main()
{
    int T;cin>>T;
    int tag=1;
    while(T--){
        cin>>n>>k;
        printf("Case %d: %lld %03lld\n",tag++,f(n),qpow(n,k));
    }
    return 0;
}

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