Description

You are given two pots, having the volume of A and B liters respectively. The following operations can be performed:

1. FILL(i)        fill the pot i (1 ≤ i ≤ 2) from the tap;
2. DROP(i)      empty the pot i to the drain;
3. POUR(i,j)    pour from pot i to pot j; after this operation either the pot j is full (and there may be some water left in the pot i), or the pot i is empty (and all its contents have been moved to the pot j).

Write a program to find the shortest possible sequence of these operations that will yield exactly C liters of water in one of the pots.

Input

On the first and only line are the numbers A, B, and C. These are all integers in the range from 1 to 100 and C≤max(A,B).

Output

The first line of the output must contain the length of the sequence of operations K. The following K lines must each describe one operation. If there are several sequences of minimal length, output any one of them. If the desired result can’t be achieved, the first and only line of the file must contain the word ‘impossible’.

Sample Input

3 5 4

Sample Output

6FILL(2)POUR(2,1)DROP(1)POUR(2,1)FILL(2)POUR(2,1)

这道题可以用宽搜做，思路挺简单的，就是写起来比较复杂，一个小错误让我改了4个小时的代码。。。

#include
    
     #include
     
      int a,b,k,num,e[111111];int vis[105][105];struct node{	int x,y,from,num,caozuo;   //from记录这个点的父亲，num表示执行了多少次，caozuo为1~6，每一个操作对应一种输出 }q[111111];int bfs(){	int front=1,rear=1,x,xx,y,yy,i,flag;	q[1].x=0;q[1].y=0;q[1].num=0;q[1].from=-1;q[1].caozuo=-1;vis[0][0]=1;	while(front<=rear)	{		x=q[front].x;y=q[front].y;		if(x==k || y==k){			num=front;			return q[front].num;		}		front++;		if(x!=a){			xx=a;yy=y;			if(vis[xx][yy]==0)			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=1;}		}		if(y!=b){			xx=x;yy=b;			if(vis[xx][yy]==0)			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=2;}		}		if(x!=0){			xx=0;yy=y;            if(vis[xx][yy]==0)			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=3;}		}		if(y!=0){	        xx=x;yy=0;            if(vis[xx][yy]==0)			{vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=4;}		}		if(x!=0){			if(y
      
       b){			    xx=x+y-b;yy=b;                if(vis[xx][yy]==0)			   {vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=5;}			 }			}		}		if(y!=0){			if(x
       
        a){			 xx=a;yy=x+y-a;             if(vis[xx][yy]==0)			  {vis[xx][yy]=1;rear++;q[rear].x=xx;q[rear].y=yy;q[rear].from=front-1;q[rear].num=q[front-1].num+1;q[rear].caozuo=6;}			 }			}		}	}	return -1;}int main(){	int i,j,t,tt;	while(scanf("%d%d%d",&a,&b,&k)!=EOF)	{		memset(vis,0,sizeof(vis));		tt=bfs();		if(tt==-1){			printf("impossible\n");continue;		}		printf("%d\n",tt);		t=0;		while(num!=1)		{			e[++t]=q[num].caozuo;			num=q[num].from;		}		for(i=t;i>=1;i--){			if(e[i]==1)printf("FILL(1)\n");			else if(e[i]==2)printf("FILL(2)\n");			else if(e[i]==3)printf("DROP(1)\n");			else if(e[i]==4)printf("DROP(2)\n");			else if(e[i]==5)printf("POUR(1,2)\n");			else if(e[i]==6)printf("POUR(2,1)\n");		}	}	return 0;}