#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

Construct a Cpp code solution to the problem outlined:

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;
}



```