
#include<bits/stdc++.h>

using namespace std;

#define b 3
#define n2 9

const int dx[4]={-1,0,1,0};
const int dy[4]={0,-1,0,1};
const char dir[4]={'u','l','d','r'};

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;
    }
};

bool istarget(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();
//        for(int i=0;i<n2;i++){
//            cout<<u.f[i]<<" ";
//        }
//        cout<<endl;
        q.pop();
        if(istarget(u)){
            return u.path;
        }
        int sx=u.space/b;
        int sy=u.space%b;
//        cout<<sx<<" "<<sy<<endl;
        for(int r=0;r<4;r++){
            int tx=sx+dx[r];
            int ty=sy+dy[r];
            if(tx<0||ty<0||tx>=b||ty>=b){
                continue;
            }
            v=u;
            v.space=tx*b+ty;
            swap(v.f[v.space],v.f[u.space]);
            if(!V[v]){
                V[v]=true;
                v.path+=dir[r];
                q.push(v);
            }
        }
    }
    return "unsolvable";
}

int main()
{
    puzzle in;
    for(int i=0;i<n2;i++){
        scanf("%d",&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;
}
/*
1 3 0
4 2 5
7 8 6
*/


### Prompt

Please create a solution in CPP to 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<bits/stdc++.h>

using namespace std;

#define b 3
#define n2 9

const int dx[4]={-1,0,1,0};
const int dy[4]={0,-1,0,1};
const char dir[4]={'u','l','d','r'};

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;
    }
};

bool istarget(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();
//        for(int i=0;i<n2;i++){
//            cout<<u.f[i]<<" ";
//        }
//        cout<<endl;
        q.pop();
        if(istarget(u)){
            return u.path;
        }
        int sx=u.space/b;
        int sy=u.space%b;
//        cout<<sx<<" "<<sy<<endl;
        for(int r=0;r<4;r++){
            int tx=sx+dx[r];
            int ty=sy+dy[r];
            if(tx<0||ty<0||tx>=b||ty>=b){
                continue;
            }
            v=u;
            v.space=tx*b+ty;
            swap(v.f[v.space],v.f[u.space]);
            if(!V[v]){
                V[v]=true;
                v.path+=dir[r];
                q.push(v);
            }
        }
    }
    return "unsolvable";
}

int main()
{
    puzzle in;
    for(int i=0;i<n2;i++){
        scanf("%d",&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;
}
/*
1 3 0
4 2 5
7 8 6
*/


```