#include <bits/stdc++.h>
using namespace std;
int n;
string s;
// 把长度为 2^n 的串 s[l]~s[r] 对应的 FBI 树的后序遍历输出
char f(int l, int r, int n)
{
    char now_res;
    if (n == 0)
    {
        if (s[l] == '0')
            now_res = 'B';
        if (s[l] == '1')
            now_res = 'I';
        cout << now_res;
        return now_res;
    }
    int len = (1 << n);
    char left_res = f(l, l + len / 2 - 1, n - 1);
    char right_res = f(l + len / 2, r, n - 1);
    if (left_res == right_res && left_res == 'B')
        now_res = 'B';
    else if (left_res == right_res && left_res == 'I')
        now_res = 'I';
    else
        now_res = 'F';
    cout << now_res;
    return now_res;
}
int main()
{
    cin >> n;
    cin >> s;
    f(0, s.size() - 1, n);
    return 0;
}

#include <bits/stdc++.h>
using namespace std;
int n;
string s;
// 把长度为 2^n 的串 s[l]~s[r] 对应的 FBI 树的后序遍历输出
void f(int l, int r, int n)
{
    if (n > 0) //有左右子树
    {
        int len = r - l + 1;
        f(l, l + len / 2 - 1, n - 1); //输出左子树的后序遍历
        f(r - len / 2 + 1, r, n - 1); //输出右子树的后序遍历
    }
    bool one, zero; //是否有 1 与 0
    one = false, zero = false;
    for (int i = l; i <= r; i++)
    {
        if (s[i] == '1')
            one = true;
        else if (s[i] == '0')
            zero = true;
    }
    //输出根节点对应的类型
    if (one == false)
        cout << 'B';
    else if (zero == false)
        cout << 'I';
    else
        cout << 'F';
}
int main()
{
    cin >> n;
    cin >> s;
    f(0, s.size() - 1, n);
    return 0;
}