/* I can do this all day */
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair < int, int > pii;
typedef pair < ll, ll > pll;
#define F first
#define S second
#define all(x) x.begin(),x.end()
#define Mp make_pair
#define point complex
#define endl '\n'
#define SZ(x) (int)x.size()
#define fast_io ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
#define file_io freopen("input.txt", "r+", stdin); freopen("output.txt", "w+", stdout);
#define mashtali return cout << "Hello, World!", 0;
const int N = 1e6 + 10;
const int LOG = 20;
const ll mod = 1e9 + 7;
const ll inf = 8e18;
const double pi = acos(-1);
const ld eps = 1e-18;
const ld one = 1.;
ll pw(ll a, ll b, ll M, ll ret = 1) { if(a == 0) return 0; a %= M; while(b) { ret = (b & 1? ret * a % M : ret), a = a * a % M, b >>= 1; } return ret % M; }
mt19937 rng(time(nullptr));
int n, A[2][N];
int dp[2][N], pd[2][N];
int main()
{
fast_io;
cin >> n;
for(int j = 0; j < 2; j ++)
{
for(int i = 1; i <= n * 2; i ++)
{
cin >> A[j][i];
}
}
A[0][0] = A[1][0] = -1e9;
for(int i = 0; i < N; i ++) for(int j = 0;j < 2; j ++) dp[j][i] = 1e9, pd[j][i] = -1e9;
dp[0][0] = dp[1][0] = 0;
pd[0][0] = pd[1][0] = 0;
for(int i = 0; i < n * 2; i ++)
{
for(int j = 0; j < 2; j ++)
{
for(int k = 0; k < 2; k ++)
{
if(A[j][i] <= A[k][i + 1])
{
dp[k][i + 1] = min(dp[k][i + 1], dp[j][i] + (k == 0));
pd[k][i + 1] = max(pd[k][i + 1], pd[j][i] + (k == 0));
}
}
}
}
/*for(int i = 1; i <= n << 1; i ++)
{
for(int j = 0; j < 2; j ++)
{
printf("i = %d j = %d dp = %d pd = %d\n", i, j, dp[j][i], pd[j][i]);
}
}
*/
int id, j;
if(dp[0][n << 1] <= n && pd[0][n << 1] >= n)
{
id = n << 1, j = 0;
}
else if(dp[1][n << 1] <= n && pd[1][n << 1] >= n)
{
id = n << 1, j = 1;
}
else
{
cout << -1 << endl;
return 0;
}
int cnt = 0;
string ans;
while(id > 0)
{
cnt += (j == 0);
if(j == 0)
{
ans += "A";
}
else
{
ans += "B";
}
///printf("id = %d j = %d dp = %d pd = %d\n", id, j, dp[j][id], pd[j][id]);
int j2 = -1;
for(int k = 0; k < 2; k ++)
{
if(A[k][id - 1] <= A[j][id] && dp[k][id - 1] <= n - cnt && pd[k][id - 1] >= n - cnt)
{
j2 = k;
}
}
if(j2 == -1)
{
assert(0);
///printf("id = %d j = %d ans = %s\n", id, j, ans.c_str());
return cout << -10, 0;
}
j = j2;
id --;
}
reverse(all(ans));
cout << ans;
return 0;
}
/* check corner case(n = 1?), watch for negetive index or overflow */
