def print_result(x, B):
    if x < 10:
        print(x)
    else:
        print('A')

M = int(input())
D = int(input())
N = int(input())
B = int(input())

x0 = M % B

# 预计算每个x的下一个状态
next_x = {}
for x in range(B):
    a = (x + D) % B
    if x == 0:
        next_x[x] = 1 % B
    else:
        next_x[x] = pow(a, x, B)

memo = {}
current_x = x0
for step in range(N + 1):
    if current_x in memo:
        # 发现循环
        first_occurrence = memo[current_x]
        cycle_length = step - first_occurrence
        remaining_steps = N - first_occurrence
        if remaining_steps <= 0:
            # 无需进入循环，直接返回当前x
            break
        else:
            # 计算剩余步骤在循环中的位置
            remainder = remaining_steps % cycle_length
            final_step = first_occurrence + remainder
            # 回溯到final_step
            temp_x = x0
            for s in range(final_step):
                temp_x = next_x[temp_x]
            current_x = temp_x
            break
    memo[current_x] = step
    if step == N:
        break
    current_x = next_x[current_x]

# 确保current_x在B范围内
current_x %= B
print_result(current_x, B)