#include<iostream>
#include<queue>
#include<map>
#include<string>

using namespace std;
#define N2 9
#define N  3
struct Puzzle
{
    int f[N2];
    int space;
    string path;
    bool operator < (const Puzzle &p)const {
            for(int i=0;i<N2;i++){
            if(f[i]==p.f[i])continue;
            return f[i]>p.f[i];
            }
        return false;
    }
};
const int dx[4]={-1,0,1,0};
const int dy[4]={0,-1,0,1};
const char dir[4]={'u','l','d','r'};
bool isTarge(Puzzle p)
{
    for(int i=0;i<N2;i++)if(p.f[i]!=i+1)return false ;
    return true;
}
string bfs(Puzzle s)
{
    queue<Puzzle> q;
    map<Puzzle,bool>V;
    Puzzle u,v;
    s.path="";
    q.push(s);
    V[s]=true;
    while(!q.empty())
    {
        u=q.front();q.pop();
        if(isTarge(u))return u.path;
        int sx=u.space/N;
        int sy=u.space%N;
        for(int i=0;i<4;i++)
        {
            int nx=sx+dx[i];
            int ny=sy+dy[i];
            if(nx<0||ny<0||nx>=N||ny>=N)continue;
            v=u;
            swap(v.f[u.space],v.f[nx*N+ny]);
            v.space=nx*N+ny;
            if(!V[v])
            {
                V[v]=true;
                v.path+=dir[i];
                q.push(v);
            }
        }
    }
    return "unsovleable";
}
int main()
{
    Puzzle in;
    for(int i=0;i<N2;i++)
    {
        cin>>in.f[i];
        if(in.f[i]==0)
        {
            in.f[i]=N2;
            in.space=i;
        }
    }
    string ans=bfs(in);
    cout<<ans.size()<<endl;
    return 0;
}


### Prompt

Create a solution in cpp for the following problem:

The goal of the 8 puzzle problem is to complete pieces on $3 \times 3$ cells where one of the cells is empty space.

In this problem, the space is represented by 0 and pieces are represented by integers from 1 to 8 as shown below.


1 3 0
4 2 5
7 8 6


You can move a piece toward the empty space at one step. Your goal is to make the pieces the following configuration in the shortest move (fewest steps).


1 2 3
4 5 6
7 8 0


Write a program which reads an initial state of the puzzle and prints the fewest steps to solve the puzzle.

Constraints

* There is a solution.

Input

The $3 \times 3$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Example

Input

1 3 0
4 2 5
7 8 6


Output

4

### Response

```cpp
#include<iostream>
#include<queue>
#include<map>
#include<string>

using namespace std;
#define N2 9
#define N  3
struct Puzzle
{
    int f[N2];
    int space;
    string path;
    bool operator < (const Puzzle &p)const {
            for(int i=0;i<N2;i++){
            if(f[i]==p.f[i])continue;
            return f[i]>p.f[i];
            }
        return false;
    }
};
const int dx[4]={-1,0,1,0};
const int dy[4]={0,-1,0,1};
const char dir[4]={'u','l','d','r'};
bool isTarge(Puzzle p)
{
    for(int i=0;i<N2;i++)if(p.f[i]!=i+1)return false ;
    return true;
}
string bfs(Puzzle s)
{
    queue<Puzzle> q;
    map<Puzzle,bool>V;
    Puzzle u,v;
    s.path="";
    q.push(s);
    V[s]=true;
    while(!q.empty())
    {
        u=q.front();q.pop();
        if(isTarge(u))return u.path;
        int sx=u.space/N;
        int sy=u.space%N;
        for(int i=0;i<4;i++)
        {
            int nx=sx+dx[i];
            int ny=sy+dy[i];
            if(nx<0||ny<0||nx>=N||ny>=N)continue;
            v=u;
            swap(v.f[u.space],v.f[nx*N+ny]);
            v.space=nx*N+ny;
            if(!V[v])
            {
                V[v]=true;
                v.path+=dir[i];
                q.push(v);
            }
        }
    }
    return "unsovleable";
}
int main()
{
    Puzzle in;
    for(int i=0;i<N2;i++)
    {
        cin>>in.f[i];
        if(in.f[i]==0)
        {
            in.f[i]=N2;
            in.space=i;
        }
    }
    string ans=bfs(in);
    cout<<ans.size()<<endl;
    return 0;
}


```