#include <iostream>
#include <cmath>]
#include <string>
#include <map>
#include <queue>
using namespace std;
#define N (3)
#define N2 (9)

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;
    }
};

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

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

int main(void){
    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

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 <iostream>
#include <cmath>]
#include <string>
#include <map>
#include <queue>
using namespace std;
#define N (3)
#define N2 (9)

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;
    }
};

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

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

int main(void){
    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;
}


```