#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
using namespace __gnu_pbds;
template<class T> using ordered_multiset =tree<T, null_type, less_equal<T>, rb_tree_tag,tree_order_statistics_node_update> ;
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file(s) freopen(s".in", "r", stdin);freopen(s".out", "w", stdout);
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define pb push_back
#define ft first
#define sd second
#define ll long long
#define pll pair<ll,ll>
const int N = 1e6 + 5;
const int M = 1e7 + 5;
const ll mod = 998244353;
const ll inf = 1e18;
ll binmul(ll x, ll ti) { ll res = 0;while (ti){if(ti & 1)res += x;x += x;ti >>= 1; x %= mod; res %= mod;} return res;}
ll binpow(ll x, ll ti) { ll res = 1;while (ti){if(ti & 1)res=binmul(res,x);x=binmul(x,x);ti >>= 1; x %= mod; res %= mod;} return res;}
ll nok(ll a, ll b) { return (a*b)/__gcd(abs(a),abs(b)) * (a*b > 0 ? 1 : -1); }
bool odd(ll n) { return (n % 2 == 1); }
bool even(ll n) { return (n % 2 == 0); }
ll n, a[N], b[N], pr[4001][2001][3];
bool dp[4001][2001][3];
const void solve() {
cin >> n;
for (int i = 1; i <= 2 * n; ++ i) cin >> a[i];
for (int i = 1; i <= 2 * n; ++ i) cin >> b[i];
dp[0][0][1] = dp[0][0][2] = 1;
for (ll i = 1; i <= 2 * n; ++ i) {
for (int j = 0; j <= min(i, n); ++ j) {
if (j > 0 && dp[i - 1][j - 1][1] && a[i] >= a[i - 1]) {
dp[i][j][1] = 1;
pr[i][j][1] = 1;
}
if (j > 0 && dp[i - 1][j - 1][2] && a[i] >= b[i - 1]) {
dp[i][j][1] = 1;
pr[i][j][1] = 2;
}
if (dp[i - 1][j][1] && b[i] >= a[i - 1]) {
dp[i][j][2] = 1;
pr[i][j][2] = 1;
}
if (dp[i - 1][j][2] && b[i] >= b[i - 1]) {
dp[i][j][2] = 1;
pr[i][j][2] = 2;
}
//cout << i << " " << j << " -> " << dp[i][j][1] << " " << dp[i][j][2] << "\n";
}
}
string ans = "";
if (!dp[n * 2][n][1] && !dp[n * 2][n][2]) {
cout << -1;
return;
}
if (dp[n * 2][n][1]) {
ll last = 1, v = n;
for (ll cur = n * 2; cur >= 1; -- cur) {
ll buf = 0;
if (last == 1) ans += "A", buf = -1;
else ans += "B";
last = pr[cur][v][last];
v += buf;
}
reverse(all(ans));
cout << ans << "\n";
} else {
ll last = 2, v = n;
for (ll cur = n * 2; cur >= 1; -- cur) {
ll buf = 0;
if (last == 1) ans += "A", buf = -1;
else ans += "B";
last = pr[cur][v][last];
v += buf;
}
reverse(all(ans));
cout << ans << "\n";
}
}
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0); cout.tie(0);
srand(time(NULL));
ll tt = 1;
//cin >> tt;
while (tt --) solve();
return 0;
}
