You are given three sequences: a_1, a_2, …, a_n; b_1, b_2, …, b_n; c_1, c_2, …, c_n.

For each i, a_i ≠ b_i, a_i ≠ c_i, b_i ≠ c_i.

Find a sequence p_1, p_2, …, p_n, that satisfy the following conditions:

  * p_i ∈ \\{a_i, b_i, c_i\}
  * p_i ≠ p_{(i mod n) + 1}.



In other words, for each element, you need to choose one of the three possible values, such that no two adjacent elements (where we consider elements i,i+1 adjacent for i<n and also elements 1 and n) will have equal value.

It can be proved that in the given constraints solution always exists. You don't need to minimize/maximize anything, you need to find any proper sequence.

Input

The first line of input contains one integer t (1 ≤ t ≤ 100): the number of test cases.

The first line of each test case contains one integer n (3 ≤ n ≤ 100): the number of elements in the given sequences.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 100).

The third line contains n integers b_1, b_2, …, b_n (1 ≤ b_i ≤ 100).

The fourth line contains n integers c_1, c_2, …, c_n (1 ≤ c_i ≤ 100).

It is guaranteed that a_i ≠ b_i, a_i ≠ c_i, b_i ≠ c_i for all i.

Output

For each test case, print n integers: p_1, p_2, …, p_n (p_i ∈ \\{a_i, b_i, c_i\}, p_i ≠ p_{i mod n + 1}).

If there are several solutions, you can print any.

Example

Input


5
3
1 1 1
2 2 2
3 3 3
4
1 2 1 2
2 1 2 1
3 4 3 4
7
1 3 3 1 1 1 1
2 4 4 3 2 2 4
4 2 2 2 4 4 2
3
1 2 1
2 3 3
3 1 2
10
1 1 1 2 2 2 3 3 3 1
2 2 2 3 3 3 1 1 1 2
3 3 3 1 1 1 2 2 2 3


Output


1 2 3
1 2 1 2
1 3 4 3 2 4 2
1 3 2
1 2 3 1 2 3 1 2 3 2

Note

In the first test case p = [1, 2, 3].

It is a correct answer, because:

  * p_1 = 1 = a_1, p_2 = 2 = b_2, p_3 = 3 = c_3 
  * p_1 ≠ p_2 , p_2 ≠ p_3 , p_3 ≠ p_1 



All possible correct answers to this test case are: [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1].

In the second test case p = [1, 2, 1, 2].

In this sequence p_1 = a_1, p_2 = a_2, p_3 = a_3, p_4 = a_4. Also we can see, that no two adjacent elements of the sequence are equal.

In the third test case p = [1, 3, 4, 3, 2, 4, 2].

In this sequence p_1 = a_1, p_2 = a_2, p_3 = b_3, p_4 = b_4, p_5 = b_5, p_6 = c_6, p_7 = c_7. Also we can see, that no two adjacent elements of the sequence are equal.
// by VIDIT BHARDWAJ
#include<bits/stdc++.h>
#include<ext/pb_ds/tree_policy.hpp>
#include<ext/pb_ds/assoc_container.hpp>

using namespace std;
using namespace __gnu_pbds;

typedef tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update> pbds;

#define int                     long long int
#define pb                      push_back
#define setbits(x)              __builtin_popcountll(x)
#define zerobits(x)             __builtin_ctzll(x)
#define ps(x,y)                 fixed<<setprecision(y)<<x // cout<<ps(1.255114,4)
#define tc(t)                   int t; cin>>t; while(t--)
#define ios                     ios_base::sync_with_stdio(false);cin.tie(NULL);
#define INTMAX                  1000001000
#define endl                    "\n"

const long long mod=1000000007;
const long long inf=1000000000000000000;
const long double pi = 3.14159265358979323846264338;

int lcm(int a,int b){
    return (a*b)/__gcd(a,b);
}
bool is_prime(int n) { 
    if (n <= 1) return false; 
    if (n <= 3) return true; 
    if (n % 2 == 0 || n % 3 == 0) return false; 
    for (int i = 5; i * i <= n; i = i + 6){
        if (n % i == 0 || n % (i + 2) == 0) return false; 
    }
    return true; 
} 
int pow(int a, unsigned int b,int mod1){
    int ans=1;
    while(b){
        if(b&1ll){
            ans*=a,ans%=mod1;
        }
        b/=2,a*=a,a%=mod1;
    }
    return ans;
}
int inverse(int a,int mod1){
    return pow(a,mod1-2,mod1);  
}
///////////////////////////////////////////////////////////////////////////////////////////////////////////

signed main(){
    ios
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);

    tc(t){
        int n; cin>>n;
        set<int> s;
        for(int i=0;i<n;i++){
            int x; cin>>x;
            s.insert(x);
        }
        for(int i=0;i<n;i++){
            int x; cin>>x;
            s.insert(x);
        }
        for(int i=0;i<n;i++){
            int x; cin>>x;
            s.insert(x);
        }
        int last=0,front=0,yo=1;
        for(int i=0;i<n-1;i++){
            if(yo){
                cout<<*(s.begin())<<" ";
                yo=0;
                front=*(s.begin());
            }
            else{
                cout<<*(s.rbegin())<<" ";
                yo=1;
                last=*(s.rbegin());
            }
        }
        
        if(n%2==0){
            if(!yo){
                cout<<*(s.rbegin())<<" ";
            }
            else{
                cout<<*(s.begin())<<" ";
            }
        }
        else{
            int num=0;
            while((*(s.begin())==last || *(s.begin())==front) && s.size()){
                s.erase(*s.begin());
            }

            cout<<*(s.begin())<<" ";
        }

        cout<<endl;
    }

    return 0;
}

/*

*/