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.
/*https://codeforces.com/profile/Amir-Kigash*/

#include<bits/stdc++.h>
using namespace std;

/*--------------Types----------------*/
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef double db;
typedef float flt;
typedef char ch;
typedef bool bl;
typedef string str;

/*--------------STLs-----------------*/
typedef pair<int, int> pi; 
typedef vector<pi> vpi;
typedef vector<int> vi;
typedef pair<ll, ll> pl;
typedef vector<pl> vpl;
typedef vector<ll> vl;

/*-----------Inputs/Outputs----------*/
#define sc1(a) scanf("%lld",&a)
#define sc2(a,b) scanf("%lld %lld",&a,&b)
#define pf1(a) printf("%lld\n",a)
#define pf2(a,b) printf("%lld %lld\n",a,b)

/*-----------Prints-Breaks-----------*/
#define sp " "
#define ndl '\n'
#define no cout<<"NO"<<ndl;
#define yes cout<<"YES"<<ndl;
#define r0 return 0;
#define brk break;
#define ctn continue;

/*--------------Loops----------------*/
#define test() ll T; cin>>T; while(T--)
#define fr0(num) for(ll i = 0; i < num; i++)
#define fr1(num) for(ll i = 1; i <= num; i++)
#define fri0(arr, num) for(ll i = 0; i < num; i++) sc1(arr[i]);
#define fri1(arr, num) for(ll i = 1; i <= num; i++) sc1(arr[i]);
#define frR(num) for(ll i = num - 1; i >= 0; i--)

/*---------------Sorts--------------*/
#define all(v) (v).begin(), (v).end()
#define rev(v) reverse(all(v))
#define srt(v) sort(all(v))
#define srtG(v) sort(all(v), greater<ll>())

/*-----------Additional-------------*/
#define pu push
#define fr front
#define bk back
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define sz size()

/*--------------STL-SC--------------*/
#define vc vector
#define lowbou lower_bound
#define upbou upper_bound

/*----------------------------------*/
#define fastread() ios_base::sync_with_stdio(false); cin.tie(NULL);

/*------------------------------GLOBAL-FUNCTIONS----------------------------------*/
const ll inf=0x3f3f3f3f;
const ll INF=1000000000;
const ll N=0;
const ll MOD=10e9+7;

void pre(ll a, ld b)
{
    cout<<fixed<<setprecision(a)<<b;
}
ll lcm(ll a, ll b)
{
    return a*b/__gcd(a, b);
}
/*------------------------------------MAIN----------------------------------------*/
signed main(/*AmirKigash*/)
{
    fastread()
    //freopen(".in", "r", stdin); freopen(".out", "w", stdout);
    test()
    {
        ll n;
        cin>>n;
        ll a[n];
        for(ll i=0; i<3; i++)
        {
            for(ll j=0; j<n; j++)
            {
                ll tm;
                if(i==0) cin>>a[j];
                else
                {
                    cin>>tm;
                    if(a[j]==a[j-1]) a[j] = tm;
                }
            }
        }
        for(ll i=0; i<n; i++) cout<<a[i]<<sp;
        cout<<ndl;
        
    }
    r0
}