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.

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).

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’.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos ( -1.0 );
const double E = 2.718281828;
typedef long long ll;

const int INF = 1000010;

using namespace std;

int n,m,x;

string s[3]= {"FILL","DROP","POUR"};
bool vis[110][110];

struct node
{
    int num,nex;///num==：1（FILL），2（DROP），3（POUR）
    int t;///倒满,guang的瓶子
    int x1,x2;///num=3时，由x1->x2
    int A,B;///a,b状态
    int step;///步数
} mp[1010];

queue<node>que;
int fro,nex;

bool bfs()
{
    while(que.size())
        que.pop();
    memset(vis,0,sizeof vis);
    node a;
    fro=0,nex=1;
    mp[fro].A=0,mp[fro].B=0;
    mp[fro].nex=-1;
    mp[fro].step=0;
    que.push(mp[fro]);
    vis[0][0]=1;
    while(que.size())
    {
        a=que.front();
        que.pop();
        if(a.A==x||a.B==x)
        {
            printf("%d\n",a.step);
            return 1;
        }
        mp[nex].nex=-1;
        if(!vis[n][a.B])///倒满A
        {
            mp[nex].nex=fro;
            mp[nex].A=n,mp[nex].B=a.B;
            mp[nex].num=1;
            mp[nex].t=1;
            mp[nex].step=a.step+1;
            que.push(mp[nex]);
            vis[n][a.B]=1;
            nex++;
        }
        if(!vis[a.A][m])///倒满B
        {
            mp[nex].nex=fro;
            mp[nex].A=a.A,mp[nex].B=m;
            mp[nex].num=1;
            mp[nex].t=2;
            mp[nex].step=a.step+1;
            que.push(mp[nex]);
            vis[a.A][m]=1;
            nex++;
        }
        if(!vis[0][a.B])///倒光A
        {
            mp[nex].nex=fro;
            mp[nex].A=0,mp[nex].B=a.B;
            mp[nex].num=2;
            mp[nex].t=1;
            mp[nex].step=a.step+1;
            que.push(mp[nex]);
            vis[0][a.B]=1;
            nex++;
        }
        if(!vis[a.A][0])///倒光B
        {
            mp[nex].nex=fro;
            mp[nex].A=a.A,mp[nex].B=0;
            mp[nex].num=2;
            mp[nex].t=2;
            mp[nex].step=a.step+1;
            que.push(mp[nex]);
            vis[a.A][m]=1;
            nex++;
        }
        if(a.A>0)///A->B
        {
            int t=m-a.B;
            if(a.A>t)//
            {
                mp[nex].A=a.A-t;
                mp[nex].B=m;
            }
            else    //
            {
                mp[nex].A=0;
                mp[nex].B=a.A+a.B;
            }
            if(!vis[mp[nex].A][mp[nex].B])
            {
                mp[nex].num=3;
                mp[nex].x1=1,mp[nex].x2=2;
                mp[nex].nex=fro;
                mp[nex].step=a.step+1;
                que.push(mp[nex]);
                vis[mp[nex].A][mp[nex].B]=1;
                nex++;
            }
        }
        if(a.B>0)///B->A
        {
            int t=n-a.A;
            if(a.B>t)//
            {
                mp[nex].A=n;
                mp[nex].B=a.B-t;
            }
            else    //
            {
                mp[nex].A=a.A+a.B;
                mp[nex].B=0;
            }
            if(!vis[mp[nex].A][mp[nex].B])
            {
                mp[nex].num=3;
                mp[nex].x1=2,mp[nex].x2=1;
                mp[nex].nex=fro;
                mp[nex].step=a.step+1;
                que.push(mp[nex]);
                vis[mp[nex].A][mp[nex].B]=1;
                nex++;
            }
        }
        fro++;
    }
    return 0;
}

void print(int x)
{
    if(mp[x].nex!=-1)
    {
        print(mp[x].nex);
        if(mp[x].num==1)
            printf("FILL(%d)\n",mp[x].t);
        else if(mp[x].num==2)
            printf("DROP(%d)\n",mp[x].t);
        else
            printf("POUR(%d,%d)\n",mp[x].x1,mp[x].x2);
    }
}

int main()
{
    while(cin>>n>>m>>x)
    {
        if(!bfs())
            cout<<"impossible"<<endl;
        else
        {
            print(fro);
        }
    }
    return 0;
}