Time Limit: 500MS

 

Memory Limit: 65536KB

 

64bit IO Format: %I64d & %I64u

 

CodeForces - 630L


Cracking the Code



Submit Status




Description




The protection of a popular program developed by one of IT City companies is organized the following way. After installation it outputs a random five digit number which should be sent in SMS to a particular phone number. In response an SMS activation code arrives.

A young hacker Vasya disassembled the program and found the algorithm that transforms the shown number into the activation code. Note: it is clear that Vasya is a law-abiding hacker, and made it for a noble purpose — to show the developer the imperfection of their protection.

The found algorithm looks the following way. At first the digits of the number are shuffled in the following order <first digit><third digit><fifth digit><fourth digit><second digit>. For example the shuffle of 12345 should lead to 13542. On the second stage the number is raised to the fifth power. The result of the shuffle and exponentiation of the number 12345 is455 422 043 125 550 171 232. The answer is the 5 last digits of this result. For the number 12345 the answer should be 71232.

Vasya is going to write a keygen program implementing this algorithm. Can you do the same?






Input




The only line of the input contains a positive integer five digit number for which the activation code should be found.






Output




Output exactly 5






Sample Input





Input



12345





Output



71232





Source



Experimental Educational Round: VolBIT Formulas Blitz



//题意：变型规则：假如给你一个五位数n为12345，将其变形为13542.（将其位置改变）。



给你一个五位数n，通过题中给的变型规则，将其变为数m，然后求出m的五次方，得到的数的后五位即为所求。



注意，若后五位有前导0,将其输出。 




#include<stdio.h>  
#include<string.h>  
#include<math.h>  
#include<algorithm>  
#include<iostream>  
#include<queue>  
#define INF 0x3f3f3f3f  
#define IN __int64  
#define ull unsigned long long  
#define ll long long  
#define N 1010  
#define M 1000000007  
using namespace std;
int main()
{
	int n,m,i,j;
	while(scanf("%d",&n)!=EOF)
	{
		int a,b,c,d,e;
		a=n/10000;
		b=n/1000%10;
		c=n/100%10;
		d=n/10%10;
		e=n%10;
		m=a*10000+c*1000+e*100+d*10+b;
		ll ans=1;
		for(i=0;i<5;i++)
			ans=(ans*m)%100000;
		printf("%05lld\n",ans);
	}
	return 0;
}