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.
// author -> LetsGoAC
#include "bits/stdc++.h"
#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using    namespace std;
using    namespace __gnu_pbds; 
#define  ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
//  ordered_set os;
//  order_of_key (k) : Number of items strictly smaller than k .
//  *find_by_order(k) : K-th element in a set (counting from zero).
typedef  long double ld;
#define  int long long
typedef  pair<int,int> pii;
#define  RAGE ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define  rep(i,n) for(i=0; i <n; i++)
#define  repv(i,k,n) for(int i=k; i<=n; i++)
#define  pb push_back
#define  mp make_pair
#define  F  first
#define  S  second
#define  sz(x)  (int)x.size()
#define  all(v) v.begin(),v.end()
#define  endl '\n'
#define  deb(x) cout<<#x<<" = "<<x<<" ";
#define  gcd   __gcd
#define  inf   LLONG_MAX
#define  minf  LLONG_MIN 
#define  gi  greater<int> 
#define  gpi greater<pii>  
#define  sp(n) fixed<<setprecision(n)
#define  nl cout<<endl
vector   <int> Fact(1000005 , 0);
const int mod = (int)(1e9+7);
int power(int x,int n)
{   
    if(n==0)    return 1;
    if(n==1)    return x%mod;
    if(n%2==0)  {   int y = power(x,n/2)%mod;return (y*y)%mod;}
    if(n&1)     {   int y = power(x,n-1);return (x%mod * y%mod)%mod;}
    return 0;
}
int modi(int n , int m)
{   //mod = m; // change const mod;
    int mod_inverse = power(n , m-2);
    return mod_inverse;
}
int  pct(int x){return __builtin_popcountll(x);}
int  lcm(int a,int b){return a*b/gcd(a,b);}
bool is_power_2(int n){ return log2(n)==floor(log2(n));}
bool is_prime(int n){if(n<=1)return 0;for(int i=2; i*i<=n; i++)if(n%i==0)return 0; return 1;}
void cal_fac()
{
     Fact[0] = Fact[1] = 1;
     for(int i=2; i<=1000000; i++)
         Fact[i] = (Fact[i-1]*i)%mod;
}
int C(int n , int k)
{
    int res = Fact[n];
    res = (res * modi( Fact[k] , mod ))%mod;
    res = (res * modi( Fact[n-k] ,mod ))%mod;
    return res;
}
ld log(ld n , int base)
{
    ld ans = 1.0 * log(n)/log(base);
    return ans;
}
void db()
{       
    #ifndef ONLINE_JUDGE
        freopen("input.txt", "r", stdin);
        freopen("output.txt", "w", stdout);
    #endif
}
void timeit()
{  
    #ifndef ONLINE_JUDGE
        cout<<"\nTime Elapsed: " <<sp(10)<< 1.0*clock() / CLOCKS_PER_SEC << " sec\n";
    #endif
}
int     dx []    =  {-1 , 0 , 1 , 0 ,-1,-1, 1,  1};
int     dy []    =  { 0 , 1 , 0 ,-1 ,-1, 1, 1, -1};
char    dir[]    =  {'U','R','D','L'};
/* ---------------------- ar.cpp --------------------------------- */
const int maxn = 1e6+5;
void solve(int tc)
{
    //  cout<<"Case #"<<tc<<": ";
    int n,i,j,k,m,l,r,q;
    cin>>n;
    int ans[n] , ar[n] , br[n] , cr[n];
    rep(i,n)
        cin>>ar[i]>>br[i]>>cr[i];
    memset(ans , -1 ,sizeof(ans));
    rep(i,n)
    {
        ans[i] = ar[i];

        if(ans[i] == ans[(i+1)%n] || ans[i] == ans[(i + n-1)%n])
            ans[i] = br[i];

        if(ans[i] == ans[(i+1)%n] || ans[i] == ans[(i + n-1)%n])
            ans[i] = cr[i];
    }
    rep(i,n)
        cout<<ans[i]<<" ";
    nl;
}
signed main()
{
    // ar.cpp
    RAGE;
    db();
    int t=1;
    cin>>t;
    for(int i=1; i<=t; i++)
    {
        solve(i);
    }
    timeit();
    return 0;
}
/*
    1. To find 2's complement , flip all the bits after 1st set bit. (right -> left
    2. continue , break;
*/  