D. Notepad

time limit per test

memory limit per test

input

output

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

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

Input

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
题目的意思就是求 ((b-1）* b ^(n-1))%c
如果用java高精度加快速幂来求，肯定会爆炸，因为有一百万位。
这道题目完全可以不用快速幂，利用同余定理就可以了，当然用了快速幂会更快一些
#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;
}