#include <bits/stdc++.h>
using namespace std;
int n;
vector <int> a, b;
// let dp[i][j][2] be the smallest end position(i)
// if we consider first i positions, j of them come from plan A
// and the last position is from plan A (0) or plan B (1)
const int maxn = 4001;
array <int, 3> goes_to[maxn][maxn][2];
bool dp[maxn][maxn][2];
bool ready[maxn][maxn][2];
bool solve(int i, int j, bool lst) {
if (i > n) {
return j == n/2;
}
if (ready[i][j][lst])
return dp[i][j][lst];
int ans = false;
// choose plan a now
if (j+1 <= n/2) {
if (lst == 0 && a[i-1] <= a[i]) {
bool res = solve(i+1, j+1, 0);
if (res) {
ans = res;
goes_to[i][j][lst] = {i+1, j+1, 0};
}
}
if (lst == 1 && b[i-1] <= a[i]) {
bool res = solve(i+1, j+1, 0);
if (res) {
ans = res;
goes_to[i][j][lst] = {i+1, j+1, 0};
}
}
}
// choose plan b now
if (lst == 0 && a[i-1] <= b[i]) {
bool res = solve(i+1, j, 1);
if (res) {
ans = res;
goes_to[i][j][lst] = {i+1, j, 1};
}
}
if (lst == 1 && b[i-1] <= b[i]) {
bool res = solve(i+1, j, 1);
if (res) {
ans = res;
goes_to[i][j][lst] = {i+1, j, 1};
}
}
dp[i][j][lst] = ans;
ready[i][j][lst] = true;
return ans;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n;
n *= 2;
a = b = vector <int> (n+1);
for (int i = 1; i <= n; i++)
cin >> a[i];
for (int i = 1; i <= n; i++)
cin >> b[i];
if (!solve(1, 0, 0)) {
cout << -1 << '\n';
return 0;
}
array <int, 3> cur = {1, 0, 0};
while (cur[0] <= n) {
array <int, 3> nxt = goes_to[cur[0]][cur[1]][cur[2]];
if (nxt[1] == cur[1] + 1)
cout << 'A';
else
cout << 'B';
cur = nxt;
}
cout << '\n';
}
