#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 Target(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(Target(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(){
    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<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 Target(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(Target(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(){
    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;
    }

```