/*
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
*/
#include<bits/stdc++.h>
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
using namespace std;
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef vector<ll>vl;
typedef vector<vl> vvl;
typedef pair<int,int>pi;
typedef pair<ll,ll> pl;
typedef vector<pl> vpl;
typedef vector<ld> vld;
typedef pair<ld,ld> pld;
typedef vector<pi> vpi;
//typedef tree<ll, null_type, less_equal<ll>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;
template<typename T> ostream& operator<<(ostream& os, vector<T>& a){os<<"[";for(int i=0; i<ll(a.size()); i++){os << a[i] << ((i!=ll(a.size()-1)?" ":""));}os << "]\n"; return os;}
#define all(x) x.begin(),x.end()
#define YES out("YES")
#define NO out("NO")
#define out(x){cout << x << "\n"; return;}
#define outfl(x){cout << x << endl;return;}
#define GLHF ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)
#define print(x){for(auto ait:x) cout << ait << " "; cout << "\n";}
#define pb push_back
#define umap unordered_map
template<typename T>
void read(vector<T>& v){
int n=v.size();
for(int i=0; i<n; i++)
cin >> v[i];
}
template<typename T>
vector<T>UNQ(vector<T>a){
vector<T>ans;
for(T t:a)
if(ans.empty() || t!=ans.back())
ans.push_back(t);
return ans;
}
void solve();
int main(){
GLHF;
int t=1;
//cin >> t;
while(t--)
solve();
}
void solve() {
int n;
cin >> n;
vvi a(2*n+1,vi(2));
for(int j=0; j<2; j++)
for(int i=1; i<=2*n; i++)
cin >> a[i][j];
vector<vvi> dp(2*n+1,vvi(2,vi(2,-1e9)));
for(int i:{0,1})for(int j:{0,1})dp[0][i][j]=0;
//dp[i][j][k] = max appearances of sequence j, at the first i terms, where the sequence k it the last used
for(int i=1; i<=2*n; i++){
for(int j=0; j<=1; j++)
for(int k=0; k<=1; k++)
for(int lst=0; lst<=1; lst++){
if(a[i][k]<a[i-1][lst])continue;
//a[i][k] >= a[i-1][lst]
if(j!=k)
dp[i][j][k] = max(dp[i][j][k], dp[i-1][j][lst]);
else
dp[i][j][k] = max(dp[i][j][k],dp[i-1][j][lst]+1);
}
}
if(*max_element(all(dp[2*n][0]))<n || *max_element(all(dp[2*n][1]))<n)out(-1)
string ans(2*n+1,'A');
int a_left=n,b_left=n;
int lst=0;
a.pb({INT_MAX,INT_MAX});
for(int i=2*n; i>=1; i--){
bool pick_a=true;
//dp[i][1][0] = max B's if I take now an A
if(dp[i][1][0]<b_left || !a_left || a[i][0]>a[i+1][lst])
pick_a = false;
if(pick_a)
ans[i]='A',a_left--,lst=0;
else
ans[i]='B',b_left--,lst=1;
}
ans=ans.substr(1,2*n);
out(ans)
}
/*
3
2 5 4 9 15 11
6 7 6 8 12 14
2
1 4 10 20
3 5 8 13
2
3 4 5 6
10 9 8 7
6
25 18 40 37 29 95 41 53 39 69 61 90
14 18 22 28 18 30 32 32 63 58 71 78
*/
