#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;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii ;
typedef pair<ll, ll> pll ;
typedef vector<pii> vii ;
typedef vector<int> veci ;
typedef vector<pll> vll ;
typedef vector<ll> vecll;
// find_by_order order_of_key
//#pragma GCC optimize("O3,unroll-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
#define F first
#define S second
#define pb push_back
#define endl '\n'
#define Mp make_pair
#define all(x) x.begin(), x.end()
#define debug(x) cerr << #x << " = " << x << endl
#define set_dec(x) cout << fixed << setprecision(x);
#define fast_io ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io freopen("in.txt" , "r" , stdin) ; freopen("out.txt" , "w" , stdout);
#define lb lower_bound
#define ub upper_bound
#define for1(n) for(int i=1;i<=n;i++)
#define for0(n) for(int i=0;i<n;i++)
#define forn(n) for(int i=n;i>0;i--)
#define pq priority_queue <pii, vector<pii>, greater<pii>>
const ll mod = 1e9+7 ;// 998244353 ;// 1e9+9;
ll inf=1e18;
const int N=1e6+100,L=21,bs=701;
int A[N],B[N],C[N],D[N],E[N],n,m,k,q,pre[N],dist[N],vis[N];
vector<int>g[N];
pii dp[N][2];
int main(){
fast_io
cin>>n;
for1(n*2){
cin>>A[i];
A[i]+=n*2-i;
}
for1(n*2){
cin>>B[i];
B[i]+=n*2-i;
}
for1(n*2){
dp[i][0]={N,0};
dp[i][1]={N,0};
pii a=dp[i-1][0],b=dp[i-1][1];
if(A[i]>=A[i-1])dp[i][0]={a.F+1,a.S+1};
if(A[i]>=B[i-1]){
dp[i][0].F=min(dp[i][0].F,b.F+1);
dp[i][0].S=max(dp[i-1][0].S,b.S+1);
}
if(B[i]>=A[i-1])dp[i][1]={a.F,a.S};
if(B[i]>=B[i-1]){
dp[i][1].F=min(dp[i][1].F,b.F);
dp[i][1].S=max(dp[i-1][1].S,b.S);
}
}
string s;
bool o=0;
if(dp[n*2][0].F<=n && dp[n*2][0].S<=n)
o=0;
else if(dp[n*2][1].F<=n && dp[n*2][1].S<=n)
o=1;
else
return cout<<"-1\n",0;
int t=n;
for(int i=n*2;i>=1;i--){
t-=1-o;
s+=('A'+o);
if(o==0){
if(A[i]>=A[i-1] && dp[i-1][0].F<=t && t<=dp[i-1][0].S)
o=0;
else
o=1;
}
else{
if(B[i]>=A[i-1] && dp[i-1][0].F<=t && t<=dp[i-1][0].S)
o=0;
else
o=1;
}
}
reverse(all(s));
cout<<s<<endl;
}
