/**
 * Author           : Dipu Kumar Mohanto (Dip)
 *                    CSE, BRUR.
 *                    Batch - 6
 *
 * Problem          : DCP-103: Make Sixovisible
 * Algorithm        : Dynamic Programing
 * Complexity       :
 * Category         : Non Classical DP
 *
 * Difficulty       : Meddium
 *
 * Source           : Dev Skill Online Judge
 *
 * Verdict          : Accepted
 *
 * Date             : 26-07-2017
 * E-mail           : dipukumarmohanto1@gmail.com
 *
 * Note             : Accepted holeo kichu question theke jay
**/


#include <bits/stdc++.h>

using namespace std;


#define  memo(a, b)             memset(a, b, sizeof(a))
#define newLine                 "\n"

#define ll                      long long


ll dp[510][450+5][2][6];   // dp[ position ][ sum of digit ][ digit taken or not ][ mod value ]

// jodi number er sob digit(500) jog kora hoy, tahole mot jogfol 4500
// tahole [ sum of digit ] ghore 4500+ howa uchit kina?
// ota dile memory limit / time limit khay
// r eta dilei accepted kemne?

int maxSum;
int tDigit;

string number;

bool solve(int pos=0, int sum=0, bool digitTaken=0, int mod=0)//, ll num=0)
{
    if (digitTaken && mod==0)
    {
        maxSum = max(maxSum, sum);
    }

    if (pos >= tDigit)
        return 0;

    ll &ret = dp[pos][sum][digitTaken][mod];

    if (ret != -1)
        return ret;

    ll ret1 = solve(pos+1, sum+(number[pos]-'0'), digitTaken|1, (mod*10+(number[pos]-'0'))%6);//, num*10+(number[pos]-'0'));
    ll ret2 = solve(pos+1, sum, digitTaken|0, mod);//, num);

    return ret = ret1 | ret2;
}

void FAST();
void TIME(double cl);

int main()
{
    //freopen("in.txt", "r", stdin);

    FAST();
    double cl = clock();

    int tc;
    cin >> tc;

    for (int tcase=1; tcase<=tc; tcase++)
    {
        cin >> number;

        tDigit = number.length();

        maxSum =  -1;
        memo(dp, -1);

        bool ans = solve();

        cout << "Case " << tcase << ": ";

        if (maxSum==-1)
            cout << "Impossible" << newLine;

        else
            cout << maxSum << newLine;
    }
    TIME(cl);
}



void FAST()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
}

void TIME(double cl)
{
    cl = clock() - cl;

    cerr << newLine << newLine << newLine
         << "------------------------------------------" << newLine
         << "Total Execution Time = " << fixed << setprecision(8)
         << (cl / CLOCKS_PER_SEC) << newLine
         << "------------------------------------------" << newLine;
}













/*****************************
         Input
5
124
3552
99912137
82113
10

         Output
Case 1: 6
Case 2: 15
Case 3: 30
Case 4: Impossible
Case 5: 0

*******************************/