#P303D. Rotatable Number
Rotatable Number
No submission language available for this problem.
Description
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.
The only line contains two space-separated integers n, x (1 ≤ n ≤ 5·106, 2 ≤ x ≤ 109).
Print a single integer — the largest b you found. If no such b exists, print -1 instead.
Input
The only line contains two space-separated integers n, x (1 ≤ n ≤ 5·106, 2 ≤ x ≤ 109).
Output
Print a single integer — the largest b you found. If no such b exists, print -1 instead.
Samples
6 11
10
5 8
-1