C++ :

#include <cstdio>
#include <cstring>
#include <algorithm>
/* 枚举 X 的所有数字排列（去重、无前导零），看是否存在 k>=2 使 perm == k*X */
int main(){
    int G; if(scanf("%d", &G) != 1) return 0;
    while(G--){
        char s[32];
        scanf("%s", s);
        int L = (int)strlen(s);
        long long X = 0;
        for(int i=0;i<L;i++) X = X * 10 + (s[i] - '0');
        std::sort(s, s + L);
        int found = 0;
        do {
            if(s[0] == '0') continue;
            long long v = 0;
            for(int i=0;i<L;i++) v = v * 10 + (s[i] - '0');
            if(v != X && v % X == 0){ found = 1; break; }
        } while(std::next_permutation(s, s + L));
        printf("%s\n", found ? "Possible" : "Impossible");
    }
    return 0;
}