1033_C. Permutation Game
After a long day, Alice and Bob decided to play a little game. The game board consists of n cells in a straight line, numbered from 1 to n, where each cell contains a number a_i between 1 and n. Furthermore, no two cells contain the same number. 

A token is placed in one of the cells. They take alternating turns of moving the token around the board, with Alice moving first. The current player can move from cell i to cell j only if the following two conditions are satisfied: 

  * the number in the new cell j must be strictly larger than the number in the old cell i (i.e. a_j > a_i), and 
  * the distance that the token travels during this turn must be a multiple of the number in the old cell (i.e. |i-j|mod a_i = 0). 



Whoever is unable to make a move, loses. For each possible starting position, determine who wins if they both play optimally. It can be shown that the game is always finite, i.e. there always is a winning strategy for one of the players.

Input

The first line contains a single integer n (1 ≤ n ≤ 10^5) — the number of numbers.

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ n). Furthermore, there are no pair of indices i ≠ j such that a_i = a_j.

Output

Print s — a string of n characters, where the i-th character represents the outcome of the game if the token is initially placed in the cell i. If Alice wins, then s_i has to be equal to "A"; otherwise, s_i has to be equal to "B". 

Examples

Input

8
3 6 5 4 2 7 1 8


Output

BAAAABAB


Input

15
3 11 2 5 10 9 7 13 15 8 4 12 6 1 14


Output

ABAAAABBBAABAAB

Note

In the first sample, if Bob puts the token on the number (not position): 

  * 1: Alice can move to any number. She can win by picking 7, from which Bob has no move. 
  * 2: Alice can move to 3 and 5. Upon moving to 5, Bob can win by moving to 8. If she chooses 3 instead, she wins, as Bob has only a move to 4, from which Alice can move to 8. 
  * 3: Alice can only move to 4, after which Bob wins by moving to 8. 
  * 4, 5, or 6: Alice wins by moving to 8. 
  * 7, 8: Alice has no move, and hence she loses immediately. 
```python
import os
import sys
from io import BytesIO, IOBase
# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# ------------------------------

from math import factorial
from collections import Counter, defaultdict
from heapq import heapify, heappop, heappush

def RL(): return map(int, sys.stdin.readline().rstrip().split())
def RLL(): return list(map(int, sys.stdin.readline().rstrip().split()))
def N(): return int(input())
def comb(n, m): return factorial(n) / (factorial(m) * factorial(n - m)) if n >= m else 0
def perm(n, m): return factorial(n) // (factorial(n - m)) if n >= m else 0
def mdis(x1, y1, x2, y2): return abs(x1 - x2) + abs(y1 - y2)
mod = 1000000007
INF = float('inf')

# ------------------------------


def main():
    n = N()
    arr = RLL()
    dic = {v: i for i, v in enumerate(arr)}
    dp = [1]*n

    for i in range(n-1, 0, -1):
        for p in range(dic[i]%i, n, i):
            if arr[p]>i and dp[dic[arr[p]]]==1:
                dp[dic[i]] = 0
                break
    print("".join(['B' if i == 1 else 'A' for i in dp]))

if __name__ == "__main__":
    main()


```