/* made by amunduzbaev */
//~ #include <ext/pb_ds/assoc_container.hpp>
//~ #include <ext/pb_ds/tree_policy.hpp>
#include "bits/stdc++.h"
using namespace std;
//~ using namespace __gnu_pbds;
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(),x.rend()
#define NeedForSpeed ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define vv vector
#define mem(arr, v) memset(arr, v, sizeof arr)
#define int long long
#define degub(x) cout<<#x<<" : "<<x<<"\n"
#define GG cout<<"here\n";
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef vector<int> vii;
typedef vector<pii> vpii;
template<class T> bool umin(T& a, const T& b) { return a > b ? a = b, true : false; }
template<class T> bool umax(T& a, const T& b) { return a < b ? a = b, true : false; }
template<int sz> using tut = array<int, sz>;
void usaco(string s) { freopen((s+".in").c_str(),"r",stdin);
freopen((s+".out").c_str(),"w",stdout); NeedForSpeed }
//~ template<class T> using oset = tree<T,
//~ null_type, less_equal<T>, rb_tree_tag,
//~ tree_order_statistics_node_update> ordered_set;
const int N = 4e3+5;
const int mod = 1e9+7;
const ll inf = 1e18;
const ld Pi = acos(-1);
#define MULTI 0
int n, m, k, s, t, q, ans, res, a[N];
int dp[N][N][2], b[N];
pii par[N][N][2];
void solve(int t_case){
cin>>n, m = 2 * n;
for(int i=1;i<=m;i++) cin>>a[i];
for(int i=1;i<=m;i++) cin>>b[i];
for(int i=0;i<=m;i++) for(int j=0;j<=n;j++) par[i][j][0] = par[i][j][1] = {-1, -1};
dp[0][0][0] = dp[0][0][1] = 1;
for(int i=1;i<=m;i++){
if(i <= n && a[i] >= a[i-1] && dp[i-1][i-1][0]) dp[i][i][0] |= 1, par[i][i][0] = {i-1, 0};
if(i <= n && b[i] >= b[i-1] && dp[i-1][0][1]) dp[i][0][1] = 1, par[i][0][1] = {0, 1};
for(int j=1;j<min(i, n + 1);j++){
// if i, j, 0 is j A's i-j B's and last is A
if(a[i] >= a[i-1] && dp[i-1][j-1][0]) dp[i][j][0] |= 1, par[i][j][0] = {j-1, 0};
if(a[i] >= b[i-1] && dp[i-1][j-1][1]) dp[i][j][0] |= 1, par[i][j][0] = {j-1, 1};
// if i, j, 1 is j A's i-j B's and last is B
if(b[i] >= a[i-1] && dp[i-1][j][0]) dp[i][j][1] |= 1, par[i][j][1] = {j, 0};
if(b[i] >= b[i-1] && dp[i-1][j][1]) dp[i][j][1] |= 1, par[i][j][1] = {j, 1};
}
}
cout<<"\n";
//~ for(int i=0;i<=n;i++){
//~ for(int j=0;j<=m;j++) cout<<dp[j][i][0]<<" ";
//~ cout<<"\n";
//~ } cout<<"\n";
//~ for(int i=0;i<=n;i++){
//~ for(int j=0;j<=m;j++) cout<<dp[j][i][1]<<" ";
//~ cout<<"\n";
//~ } cout<<"\n";
int i = m, j = n, t = 0;
if(dp[m][n][0]) t = 0;
else if(dp[m][n][1]) t = 1;
else { cout<<"-1\n"; return; }
vii tt;
while(i){ tt.pb(t);
pii pp = par[i][j][t];
j = pp.ff, t = pp.ss, i--;
}
reverse(all(tt));
for(auto x : tt){
if(x) cout<<"B";
else cout<<"A";
} cout<<"\n";
}
signed main(){
NeedForSpeed
if(MULTI){
int t; cin>>t;
for(int t_case = 1; t_case <= t; t_case++) solve(t_case);
} else solve(1);
return 0;
}
