Adjacent array elements not divisible by 3

/** An array of integers is provided.
You need to arrange the numbers such that no 2 consecutive numbers sum is divisible by 3.
If no such arrangement is possible then return(“impossible”), else return the arrangement.
**/

#include<bits/stdc++.h>
#define pb push_back

using namespace std;

int T, N, x, z, o, t;
vector<int> zeros;
vector<int> ones;
vector<int> twos;
list<int> ans;

// Input:
// 7
// 11
// 1 1 0 0 0 0 0 0 2 2 2
// 9
// 1 1 0 0 0 0 0 0 2
// 6
// 0 1 0 1 2 2
// 4
// 0 0 0 1
// 4
// 1 2 1 2
// 4
// 2 2 2 1
// 5
// 0 0 0 2 2

// Output:
// 0 1 0 1 0 2 0 2 0 2 0
// Impossible
// 0 1 1 0 2 2
// Impossible
// Impossible
// Impossible
// 0 2 0 2 0


int solve(){
    z = zeros.size();
    o = ones.size();
    t = twos.size();
    if(!z && o && t)    // if no 0, but both 1 and 2
        return 0;
    if(z && !o && !t)   // if only 0
        return 0;
    if(o+t+1 < z)       // if more 0s than total+1 :: there can be maximum total+1 places for 0s for valid arrangement
        return 0;
    return 1;
}

void arrange(){
    if(o && t)          // if both 1s and 2s are there, keep aside a 0 to be kept in between the 1's series and 2's series
        z--;
    for(int i=0; i<o; ++i){     // fill up 0s and 1s
        if(z)
            ans.pb(zeros[--z]);
        ans.pb(ones[i]);
    }
    if(o && t)          // if both 1s and 2s are there, the kept aside 0 needs to be put now
        z++;
    for(int i=0; i<t; ++i){     // fill up 0s and 2s
        if(z)
            ans.pb(zeros[--z]);
        ans.pb(twos[i]);
    }
    if(z)               // if there's a last zero left
        ans.pb(zeros[--z]);
    for(auto it=ans.begin(); it!=ans.end(); ++it)
        cout<<*it<<" ";
    cout<<endl;
}

int main(){
    cin>>T;
    while(T--){
        cin>>N;
        zeros.clear();
        ones.clear();
        twos.clear();
        ans.clear();
        for(int i=0; i<N; ++i){
            cin>>x;
            if(!(x%3))
                zeros.pb(x);
            else if(x%3 == 1)
                ones.pb(x);
            else
                twos.pb(x);
        }
        int res = solve();
        if(!res)
            cout<<"Impossible"<<endl;
        else
            arrange();
    }
}