Ekka and his friend Dokka decided to buy a cake. They both love cakes and that’s why they want to share the cake after buying it. As the name suggested that Ekka is very fond of odd numbers and Dokka is very fond of even numbers, they want to divide the cake such that Ekka gets a share of N square centimeters and Dokka gets a share of M square centimeters where N is odd and M is even. Both N and M are positive integers.

They want to divide the cake such that N * M = W, where W is the dashing factor set by them. Now you know their dashing factor, you have to find whether they can buy the desired cake or not.

Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains an integer W (2 ≤ W < 263). And W will not be a power of 2.

Output
For each case, print the case number first. After that print “Impossible” if they can’t buy their desired cake. If they can buy such a cake, you have to print N and M. If there are multiple solutions, then print the result where M is as small as possible.

Sample Input
3
10
5
12
Sample Output
Case 1: 5 2
Case 2: Impossible
Case 3: 3 4

題意： 給一個數字，問滿足 n * m = w ，其中n為偶數，m為奇數。
現在求最小的滿足式子的n。
分析： 我們可以從算數基本定理來考慮，對於一個數字，如果其是偶數的話，其質因子分解是什麼樣子？ 那麼其質因子2的指數一定是>=1,所以我們現在想求最小的偶數 * 奇數， 其實就是求最大奇數，最大的奇數不就是將所有的質因子2都去掉不就是完全奇數了。
所以就很明確了。
程式碼

#include<bits/stdc++.h>
using namespace std;
typedef pair<int,int>pii;
#define first fi
#define second se
#define  LL long long
#define fread() freopen("in.txt","r",stdin)
#define fwrite() freopen("out.txt","w",stdout)
#define CLOSE() ios_base::sync_with_stdio(false)

const int MAXN = 1e5;
const int MAXM = 1e6;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;

int main(){
    CLOSE();
//  fread();
//  fwrite();
    int T ;scanf("%d",&T);int ncase=1;
    while(T--){
        printf("Case %d: ",ncase++);
         LL  w,m;scanf("%lld",&w);
         if(w&1){
            puts("Impossible");
            continue;
         }
         int cnt=0; m=1;
         while(w){
            if(w&1) break;
            cnt++;
            w>>=1;
         } 
         //printf("cnt==%d\n",cnt);
         printf("%lld %lld\n",w,(m<<cnt)); 
    } 
    return 0;
}