CF303D.Rotatable Number

Bike is a smart boy who loves math very much. He invented a number called "Rotatable Number" inspired by 142857.

As you can see, 142857 is a magic number because any of its rotatings can be got by multiplying that number by 1, 2, ..., 6 (numbers from one to number's length). Rotating a number means putting its last several digit into first. For example, by rotating number 12345 you can obtain any numbers: 12345, 51234, 45123, 34512, 23451. It's worth mentioning that leading-zeroes are allowed. So both 4500123 and 0123450 can be obtained by rotating 0012345. You can see why 142857 satisfies the condition. All of the 6 equations are under base 10.

• 142857·1 = 142857;
• 142857·2 = 285714;
• 142857·3 = 428571;
• 142857·4 = 571428;
• 142857·5 = 714285;
• 142857·6 = 857142.

Now, Bike has a problem. He extends "Rotatable Number" under any base b. As is mentioned above, 142857 is a "Rotatable Number" under base 10. Another example is 0011 under base 2. All of the 4 equations are under base 2.

• 0011·1 = 0011;
• 0011·10 = 0110;
• 0011·11 = 1001;
• 0011·100 = 1100.

So, he wants to find the largest b (1 < b < x) so that there is a positive "Rotatable Number" (leading-zeroes allowed) of length n under base b.

Note that any time you multiply a rotatable number by numbers from 1 to its length you should get a rotating of that number.

Bike 是一个聪明的男孩，非常热爱数学。他受数字 142857 的启发，发明了一类被称为“可旋转数”（Rotatable Number）的数。

如你所见，142857 是一个神奇的数，因为它的任意一种旋转形式均可通过将该数分别乘以 $1, 2, \dots, 6$（即从 $1$ 到该数的位数）得到。“旋转一个数”是指将该数末尾若干位数字移到最前面。例如，对数字 $12345$ 进行旋转，可得到如下所有数字：$12345,\ 51234,\ 45123,\ 34512,\ 23451$。需要特别说明的是，允许前导零。因此，$4500123$ 和 $0123450$ 均可通过旋转 $0012345$ 得到。你可以由此理解为何 $142857$ 满足该条件。以下全部等式均在十进制（base 10）下成立：

• $142857 \cdot 1 = 142857$；
• $142857 \cdot 2 = 285714$；
• $142857 \cdot 3 = 428571$；
• $142857 \cdot 4 = 571428$；
• $142857 \cdot 5 = 714285$；
• $142857 \cdot 6 = 857142$。

现在，Bike 遇到了一个问题：他将“可旋转数”的定义推广到了任意进制 $b$。如前所述，$142857$ 是一个十进制下的“可旋转数”。另一个例子是二进制（base 2）下的 $0011$。以下全部等式均在二进制下成立：

• $0011 \cdot 1 = 0011$；
• $0011 \cdot 10 = 0110$；
• $0011 \cdot 11 = 1001$；
• $0011 \cdot 100 = 1100$。

于是，他希望找出满足以下条件的最大进制 $b$（其中 $1 < b < x$）：存在一个长度为 $n$ 的正整数“可旋转数”（允许前导零），且该数在进制 $b$ 下是可旋转的。

注意：每次将该可旋转数乘以从 $1$ 到其位数（即 $n$）之间的任意整数时，所得结果必须恰好是该数的某一种旋转形式。

The only line contains two space-separated integers n, x (1 ≤ n ≤ 5·106, 2 ≤ x ≤ 109).

唯一一行包含两个以空格分隔的整数 $n$ 和 $x$（$1 \leq n \leq 5\cdot10^6$，$2 \leq x \leq 10^9$）。

Print a single integer — the largest b you found. If no such b exists, print -1 instead.

输出一个整数——即你找到的最大的 b。如果不存在这样的 b，则输出 -1。