Chinese Rings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 261 Accepted Submission(s): 157

Problem Description

Dumbear likes to play the Chinese Rings (Baguenaudier). It’s a game played with nine rings on a bar. The rules of this game are very simple: At first, the nine rings are all on the bar.
The first ring can be taken off or taken on with one step.
If the first k rings are all off and the (k + 1)th ring is on, then the (k + 2)th ring can be taken off or taken on with one step. (0 ≤ k ≤ 7)

Now consider a game with N (N ≤ 1,000,000,000) rings on a bar, Dumbear wants to make all the rings off the bar with least steps. But Dumbear is very dumb, so he wants you to help him.

Input

Each line of the input file contains a number N indicates the number of the rings on the bar. The last line of the input file contains a number "0".

Output

For each line, output an integer S indicates the least steps. For the integers may be very large, output S mod 200907.

Sample Input

1
4
0

Sample Output

1
10

//推的過程：
//f(n)代表把n個環全部拿下來要用的次數
//要將第n個拿下， 前n-2個要拿下來，所以要用f(n-2)+1，並且第n-1個在環上,要將第n-1個拿下來，必//須前n-2個環要放回去即f(n-2)，然後再拿下前n-3，然後纔拿下第n-1個,即（f(n-3)+1）,然後
//再把前//n-3個放回去，。。。一直循環得到: f(n)=f(n-2)+1+f(n-2)+f(n-3)+1+f(n-3)+......+f(n-n);  
 
//於是遞推公式爲:f(n)=2*f(n-2)+f(n-1)+1;  顯然是構造二分矩陣求解

//或者在拿第n個時要用f(n-2)*2+1個，乘以2是拿下第n個後要將前n-2個放回去，接着你想想拿完第

//個後要拿下剩下的n-1個就要用到剛纔的定義——》第二句，要用f(n-1)次，所以總的次數是：

//f(n)=f(n-2)*2+f(n-1)+1 
//構造矩陣爲
// 1  2  1   f(2)=2
// 1  0  0   f(1)=1
// 0  0  0    1

#include<stdio.h>
struct node
{
	long long num[3][3];//注意這裏要用long long 或__int64
};
node res,tp;
node mut(node a,node b)
{
	node c;
	int i,j,k;
	for(i=0;i<3;i++)
		for(j=0;j<3;j++)
		{
			c.num[i][j]=0;
			for(k=0;k<3;k++)
				c.num[i][j]+=(a.num[i][k]*b.num[k][j])%200907;
			c.num[i][j]%=200907;
		}
		return c;
}
node abc(int temp,node p)
{
	tp=p;
	p=res;
	while(temp)
	{
		if(temp&1)
			p=mut(p,tp);
		tp=mut(tp,tp);
		temp>>=1;
	}
	return p;
}
int main()
{
	int n;
	res.num[0][0]=res.num[1][1]=res.num[2][2]=1;
	res.num[0][1]=res.num[0][2]=res.num[2][0]=0;
	res.num[2][1]=res.num[1][0]=res.num[1][2]=0;
	node p;
	p.num[0][0]=p.num[0][2]=p.num[1][0]=p.num[2][2]=1;
	p.num[1][1]=p.num[1][2]=p.num[2][0]=p.num[2][1]=0;
	p.num[0][1]=2;
	while(scanf("%d",&n),n)
	{
		node k;
		if(n<3)
		{
			printf("%d/n",n);
			continue;
		}
		k=abc(n-2,p);
		int sum=(k.num[0][0]*2+k.num[0][1]+k.num[0][2])%200907;
		printf("%d/n",sum);
	}
	return 0;
}