D. Notepad time limit per test 2 seconds memory limit per test 64 megabytes input standard input output standard output

Nick is attracted by everything unconventional. He doesn't like decimal number system any more, and he decided to study other number systems. A number system with base b caught his attention. Before he starts studying it, he wants to write in his notepad all the numbers of length n without leading zeros in this number system. Each page in Nick's notepad has enough space for c

 numbers exactly. Nick writes every suitable number only once, starting with the first clean page and leaving no clean spaces. Nick never writes number 0 as he has unpleasant memories about zero divide.

Would you help Nick find out how many numbers will be written on the last page.

Input

The only input line contains three space-separated integers b

, n and c (2 ≤ b < 10106, 1 ≤ n < 10106, 1 ≤ c ≤ 109). You may consider that Nick has infinite patience, endless amount of paper and representations of digits as characters. The numbers doesn't contain leading zeros.

Output

In the only line output the amount of numbers written on the same page as the last number.

Examples input

2 3 3

output

1

input

2 3 4

output

4

題目的意思就是求 ((b-1）* b ^(n-1))%c

如果用java高精度加快速冪來求，肯定會爆炸，因為有一百萬位。

這道題目完全可以不用快速冪，利用同餘定理就可以了，當然用了快速冪會更快一些

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
#include <queue>

using namespace std;
#define MAX 1000000
char b[MAX+5];
char n[MAX+5];
long long int c;
int main()
{
    while(scanf("%s%s%lld",b,n,&c)!=EOF)
    {


    long long int num=0;//b
    long long int num2=0;//b-1
    int len=strlen(b);
    //b
    for(int i=0;i<len;i++)
        num=(b[i]-'0'+num*10)%c;
    //b-1
    for(int i=len-1;i>=0;i--)
      if(b[i]!='0'){b[i]--;break;}
      else  b[i]='9';
    for(int i=0;i<len;i++)
        num2=(b[i]-'0'+num2*10)%c;
        
    int len2=strlen(n);
    long long int ans=0;
    long long int num3=num;
    //n-1
    for(int i=len2-1;i>=0;i--)
      if(n[i]!='0'){n[i]--;break;}
      else  n[i]='9';
     
    for(int j=0;j<len2;j++)
    {
        if(n[j]!='0')
        {
            if(j!=0)
            {
                 long long int num4=num;
                 for(int k=1;k<=9;k++)
                     num=((num%c)*num4)%c;
            }
            int x=n[j]-'0';
            if(j==0)
                x--;
            for(int p=1;p<=x;p++)
              num=((num%c)*num3)%c;
        }
        else
        {
            if(j!=0)
            {
                long long int num4=num;
                for(int k=1;k<=9;k++)
                    num=((num%c)*num4)%c;
            }
            else
               num=1;
        }
    }
    num=((num%c)*(num2%c))%c;
    if(num==0)
        num=c;
    printf("%lld\n",num);
    }
    return 0;
}