#include <iostream>
using namespace std;
/*
Let us sort all the 4N coloring plans together, first by height then by building index.
Out of this sequence, we have to choose 2N elements such that:
No two coloring plans for a single building are chosen.
*/
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int N;
cin >> N;
int A[2*N + 1];
int B[2*N + 1];
for(int i = 1; i <= 2*N; i++) cin >> A[i];
for(int i = 1; i <= 2*N; i++) cin >> B[i];
A[0] = B[0] = 0;
int lo[2*N + 1][2], hi[2*N + 1][2]; //lower & upper limits on number of A buildings.
lo[0][0] = hi[0][0] = 0;
lo[0][1] = hi[0][1] = 0;
for(int i = 1; i <= 2*N; i++)
{
lo[i][0] = lo[i][1] = 1e9;
hi[i][0] = hi[i][1] = -1e9;
if(A[i-1] <= A[i])
{
hi[i][0] = max(hi[i][0], hi[i-1][0] + 1);
lo[i][0] = min(lo[i][0], lo[i-1][0] + 1);
}
if(B[i-1] <= A[i])
{
hi[i][0] = max(hi[i][0], hi[i-1][1] + 1);
lo[i][0] = min(lo[i][0], lo[i-1][1] + 1);
}
if(A[i-1] <= B[i])
{
hi[i][1] = max(hi[i][1], hi[i-1][0]);
lo[i][1] = min(lo[i][1], lo[i-1][0]);
}
if(B[i-1] <= B[i])
{
hi[i][1] = max(hi[i][1], hi[i-1][1]);
lo[i][1] = min(lo[i][1], lo[i-1][1]);
}
// cerr << "i = " << i << '\n';
// cerr << lo[i][0] << ' ' << hi[i][0] << " | " << lo[i][1] << ' ' << hi[i][1] << '\n';
}
if(!(lo[2*N][0] <= N && N <= hi[2*N][0]) && !(lo[2*N][1] <= N && N <= hi[2*N][1]))
{
cout << "-1\n";
return 0;
}
string res(2*N, '.');
int i = 2*N;
int lb = (lo[i][1] <= N && N <= hi[i][1]);
int buildA = N;
for(int i = 2*N; i >= 1; i--)
{
if(lb == 0)
{
res[i-1] = 'A';
buildA--;
}
else res[i-1] = 'B';
if(lb == 0)
{
if(A[i-1] <= A[i] && lo[i-1][0] <= buildA && buildA <= hi[i-1][0])
lb = 0;
else
lb = 1;
}
else
{
if(A[i-1] <= B[i] && lo[i-1][0] <= buildA && buildA <= hi[i-1][0])
lb = 0;
else
lb = 1;
}
}
cout << res << '\n';
}
