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:kkdrummer----/
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
//#include <boost/multiprecision/cpp_int.hpp> 
//using namespace boost::multiprecision; 
using namespace __gnu_pbds;
using namespace std;
typedef    long long                	 ll;
typedef    long double              	 ld;
typedef    unordered_set<ll>         	 usll;
typedef    unordered_multiset<ll>    	 umsll;
typedef    multiset<ll>              	 msll;
typedef    set<ll>                  	 sll;
typedef    vector<ll>               	 vll;
typedef    pair<ll,ll>              	 pll;
typedef    vector<pll>              	 vpll;
typedef    priority_queue<ll>       	 pqll;
typedef    vector<int>              	 vi;
typedef    set<int>                 	 si;
typedef    multiset<int>            	 msi;
typedef    unordered_multiset<int>  	 umsi;
typedef    unordered_set<int>       	 usi;
typedef    pair<int,int>               	 pi;
typedef    vector<pi>                	 vpi;
typedef    priority_queue<int>       	 pqi;
typedef    tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>   ind_si;
typedef    tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>   ind_sll;
typedef    tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>   ind_msi;
typedef    tree<ll,null_type,less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update>   ind_msll;
#define    in          		insert
#define    fi          		first
#define    se          		second
#define    pb   	    	push_back
#define    mp   	   		make_pair
#define    be   	   		begin
#define    en   	   		end
#define    itr  	   		iterator
#define    io  	       		ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define    mo   	   		1000000007
#define    inf  	   		8222372026854775807
#define    ninf   	   		-inf
#define    ima   	   		1147483647
#define    imi   	   		-ima
#define    oncnt       		__builtin_popcount
#define    zerobegin   		__builtin_clz
#define    zeroend    		__builtin_ctz
#define    parity      		__builtin_parity
#define    all(x)      		x.be(),x.en()
#define    eps         		1e-9
#define    coutd       		cout<<setprecision(10)<<fixed
#define    mems(dp,x)  		memset(dp,x,sizeof(dp))
#define    fbo         		find_by_order
#define    ook    	   		order_of_key
#define    upb    	   		upper_bound
#define    lowb  	   		lower_bound 
#define    lte(v,x)    		(upb(all(v),x)-v.be())
#define    gte(v,x)    		(v.end()-lowb(all(v),x))
#define    gt(v,x)     		(v.en()-upb(all(v),x))
#define    lt(v,x)     		(lowb(all(v),x)-v.be())
const      ld PI= 			3.1415926535897932384626433832792884197169399375105820974944;
inline ll modpow(ll x,ll n){if(n==0)return 1;if(x==0)return 0;if(n==1)return(x%mo);ll u=(modpow(x,n/2));u=(u*u)%mo;if(n%2!=0)u=(u*x%mo)%mo;return u;}
inline ll modinv(ll x){return modpow(x,mo-2);}
inline ll mmul(ll a,ll b){if(a>=mo)a=a%mo;if(b>=mo)b=b%mo;if(a*b>=mo)return(a*b)%mo;return(a*b);}
inline ll madd(ll a, ll b){if(a>=mo)a=a%mo;if(b>=mo)b=b%mo;if(a+b>=mo)return(a+b)%mo;return(a+b);}
inline ll msub(ll a, ll b){if(a>=mo)a=a%mo;if(b>=mo)b=b%mo;return(((a-b)%mo+mo)%mo);}
inline ll mdiv(ll a,ll bb){if(a>=mo)a=a%mo;ll b=modinv(bb);if(b>=mo)b=b%mo;if(a*b>=mo)return(a*b)%mo;return(a*b);}
inline ll gcd(ll a,ll b){return __gcd(a,b);}
inline ll lcm(ll a,ll b){return (a*b)/gcd(a,b);}
 






int main()
{               io 
int testcases=1;                       cin>>testcases;
while(testcases--)
{

int n;
cin>>n;
vi v;
int a[n],b[n],c[n];
for(int i=0;i<n;i++)cin>>a[i];
for(int i=0;i<n;i++)cin>>b[i];
for(int i=0;i<n;i++)cin>>c[i];

v.pb(a[0]);
for(int i=1;i<n;i++)
{
	if(a[i]!=v.back())
	v.pb(a[i]);
	else if(b[i]!=v.back())
	v.pb(b[i]);
	else
	v.pb(c[i]);
}
for(int i=0;i<n;i++)cout<<v[i]<<" ";
cout<<"\n";






















































































































































































}return 0;}
