#include<iostream>
#include<cmath>
#include<string>
#include<map>
#include<queue>
#define N 3
#define N2 9
using namespace std;
 
struct Puz{
  int f[N2];
  int space;
  string path;
  
  bool operator < (const Puz &p) const{
    for(int i = 0; i < N2; i++){
      if(f[i]==p.f[i]) continue;
      
      return f[i]>p.f[i];
      
    }
    
    return false;
    
  }
};
 
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
char dir[4]={'u','l','d','r'};
 
bool check(Puz p){
  for(int i = 0; i < N2; i++){
    if(p.f[i] != (i+1)) return false;
  }
  
  return true;

}
 
string bfs(Puz s){
  queue<Puz> Q;
  map<Puz,bool> V;
  Puz u, v;
  
  s.path="";
  Q.push(s);
  V[s]=true;
   
  while(!Q.empty()){
    u = Q.front();
    Q.pop();
    
    if(check(u)) return u.path;
    
    int sx = u.space / N;
    int sy = u.space % N;
    
    for(int i = 0; i < 4; i++){
      int tx = sx + dx[i];
      int ty = sy + dy[i];
      
      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[i];
	Q.push(v);
      }
    }
  }
  
  return "unsolvable";
  
}
 
int main(){
  Puz 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>
#define N 3
#define N2 9
using namespace std;
 
struct Puz{
  int f[N2];
  int space;
  string path;
  
  bool operator < (const Puz &p) const{
    for(int i = 0; i < N2; i++){
      if(f[i]==p.f[i]) continue;
      
      return f[i]>p.f[i];
      
    }
    
    return false;
    
  }
};
 
int dx[4]={-1,0,1,0};
int dy[4]={0,-1,0,1};
char dir[4]={'u','l','d','r'};
 
bool check(Puz p){
  for(int i = 0; i < N2; i++){
    if(p.f[i] != (i+1)) return false;
  }
  
  return true;

}
 
string bfs(Puz s){
  queue<Puz> Q;
  map<Puz,bool> V;
  Puz u, v;
  
  s.path="";
  Q.push(s);
  V[s]=true;
   
  while(!Q.empty()){
    u = Q.front();
    Q.pop();
    
    if(check(u)) return u.path;
    
    int sx = u.space / N;
    int sy = u.space % N;
    
    for(int i = 0; i < 4; i++){
      int tx = sx + dx[i];
      int ty = sy + dy[i];
      
      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[i];
	Q.push(v);
      }
    }
  }
  
  return "unsolvable";
  
}
 
int main(){
  Puz 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;
}


```