An Easy Problem!

Description

Have you heard the fact "The base of every normal number system is 10" ? Of course, I am not talking about number systems like Stern Brockot Number System. This problem has nothing to do with this fact but may have some similarity. 

You will be given an N based integer number R and you are given the guaranty that R is divisible by (N-1). You will have to print the smallest possible value for N. The range for N is 2 <= N <= 62 and the digit symbols for 62 based number is (0..9 and A..Z and a..z). Similarly, the digit symbols for 61 based number system is (0..9 and A..Z and a..y) and so on.

Input

Each line in the input will contain an integer (as defined in mathematics) number of any integer base (2..62). You will have to determine what is the smallest possible base of that number for the given conditions. No invalid number will be given as input. The largest size of the input file will be 32KB.

Output

If number with such condition is not possible output the line "such number is impossible!" For each line of input there will be only a single line of output. The output will always be in decimal number system.

Sample Input

3
5
A

Sample Output

4
6
11

题意：给定一个N进制的数，由0-9、a-z、A-Z组成，最大62进制。已知这个数能被N-1整除。问满足条件的N至少是多少。
思路：先对输入字符进行判定，进制数肯定大于字符代表的数。例如，出现0-9，进制肯定是10以上。而对于整除的判定操作，其实有一种很巧妙的判定方式——数字和的整除性。一个N进制数abc=a*n^2+b*n^2+c。abc%(n-1)=0可推出a*n^2%(n-1)+b*n^2%(n-1)+c%(n-1)=0。n%(n-1)=1。所以化简后就是(a+b+c)%(n-1)=0。

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long LL;
int w(char c)
{
    if(c<='9')
        return c-'0';
    if(c<='Z')
        return c-'A'+10;
    return c-'a'+36;
}

int main()
{
    int i,t,sum,max;
    char c;
    while((c=getchar())>0)
    {
        sum=w(c);
        t=sum;
        max=t;
        while((c=getchar())!='\n')
        {
            t=w(c);
            if(t>max)
                max=t;
            sum+=t;
        }
        for(i=max; i<62; i++)
        {
            if(sum%i==0)
            {
                printf("%d\n",i+1);
                break;
            }
        }
        if(i==62)
            printf("such number is impossible!\n");
    }
    return 0;
}