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.
// #pragma comment(linker, "/stack:200000000")
// #pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

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

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

//========================TypeDefs===================================

typedef long long int ll;
typedef unsigned long long int ull;
typedef long double ldb;

typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

//=========================MACROS====================================

#define Shen() main()
#define pb push_back
#define popb pop_back()
#define pf push_front
#define popf pop_front()
#define si size()
#define be begin()
#define en end()
#define all(v) v.be, v.en
#define le length()
#define mp make_pair
#define mt make_tuple
#define acc(v) accumulate(all(v), 0)
#define fi first
#define se second

#define vi vector<int>
#define vll vector<ll>
#define vpll vector<pair<ll,ll>>
#define vvi vector<vector<int>>
#define vvll vector<vll>

#define mli map<ll,int>
#define mll map<ll,ll>

#define forz(i, n) for (int i = 0; i < n; i++)
#define fora(i, m, n) for (int i = m; i < n; i++)
#define rforz(i, n) for (int i = n - 1; i >= 0; i--)
#define rfora(i, n, m) for (int i = n; i >= m; i--)

#define deci(n) fixed << setprecision(n)
#define high(n) __builtin_popcount(n)
#define highll(n) __builtin_popcountll(n)
#define parity(n) __builtin_parity(n)
#define clz(n) __builtin_clz(n)
#define clzll(n) __builtin_clzll(n)
#define ctz(n) __builtin_ctz(n)

#define bset(a, p) ((a) | (1ll << (p)))
#define bchk(a, p) ((a) & (1ll << (p)))
#define bxor(a, p) ((a) ^ (1ll << (p)));
#define brem(a, p) (bchk(a, p) ? (bxor(a, p)) : (a))

#define lb lower_bound
#define ub upper_bound
#define er equal_range

#define findnot find_first_not_of

#define maxe *max_element
#define mine *min_element

#define MOD 1000000007
#define MOD2 998244353
#define PI 3.1415926535897932384
#define INF LLONG_MAX

#define gcd __gcd
#define fastio ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)

#define endl "\n"
#define P0(a) cout << a << " "
#define P1(a) cout << a << endl
#define P2(a, b) cout << a << " " << b << endl
#define P3(a, b, c) cout << a << " " << b << " " << c << endl
#define P4(a, b, c, d) cout << a << " " << b << " " << c << " " << d << endl

#define oset tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update>
#define osetll tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>

#define ofk order_of_key
#define fbo find_by_order
//member functions :
//1. order_of_key(k) : number of elements sbtriectly lesser than k
//2. find_by_order(k) : k-th element in the set

//==============================FUNCTIONS===========================================

auto clk = clock();
mt19937_64 rang(chrono::high_resolution_clock::now().time_since_epoch().count());
void run_time()
{
    cout << endl;
    cout << "Time elapsed: " << (double)(clock() - clk) / CLOCKS_PER_SEC << endl;
}

inline ll power(ll x, ll y, ll p = MOD)
{
    ll res = 1;
    x = x % p;
    while (y > 0)
    {
        if (y & 1)
            res = (res * x) % p;
        y = y >> 1;
        x = (x * x) % p;
    }
    return res;
}

inline ll modadd(ll a, ll b, ll m = MOD)
{
    a += b;
    if (a >= m)
        a -= m;
    return a;
}

inline ll modmul(ll a, ll b, ll m = MOD)
{
    return ((a % m) * (b % m)) % m;
}

inline ll modi(ll a, ll m = MOD) { return power(a, m - 2, m); }

long long int inverse(long long int i,ll m=MOD){
    if(i==1) return 1;
    return (m - ((m/i)*inverse(m%i,m))%m+m)%m;
}

//================================CODE=============================================

int Shen()
{
    fastio;
    ll tcase;
    cin >> tcase;
    while (tcase--)
    {
        ll n;
        cin>>n;
        vll a(n),b(n),c(n),p(n);
        
        fora(i,0,n)
        cin>>a[i];
        fora(i,0,n)
        cin>>b[i];
        fora(i,0,n)
        cin>>c[i];
        p[0]=a[0];
        ll prev = p[0];
        fora(i,1,n)
        {
            if(p[i-1]!=a[i])
            p[i] = a[i];
            else
            {
                p[i]=b[i];
            }
            
        }
        fora(i,0,n)
        cout<<p[i]<<" ";
        cout<<endl;
    }
    //run_time();
    return 0;
}