#include <iostream>
#include <cmath>
#include <string>
#include <map>
#include <queue>
using namespace std;

#define A 3
#define AB 9

struct pazzle{
  int f[AB];
  int space;
  string path;

  bool operator < (const pazzle &p) const{
    for(int i=0;i<AB;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(pazzle p){
  for(int i=0;i<AB;i++)
    if(p.f[i]!=(i+1))return false;
  return true;
}

string bfs(pazzle s){
  queue<pazzle> Q;
  map<pazzle,bool> V;
  pazzle 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/A;
    int sy = u.space%A;

    for(int r=0;r<4;r++){
      int tx = sx+dx[r];
      int ty = sy+dy[r];

      if(tx<0 || ty<0 || tx>=A || ty>=A)continue;

      v=u;
      swap(v.f[u.space],v.f[tx*A+ty]);
      v.space = tx*A+ty;
      if(!V[v]){
        V[v]=true;
        v.path += dir[r];
        Q.push(v);
      }
    }
  }
  return "unsolvable";
}

int main(){
  pazzle in;

  for(int i=0;i<AB;i++){
    cin >> in.f[i];
    if(in.f[i]==0){
      in.f[i]=AB;
      in.space =i;
    }
  }
 string ans = bfs(in);
 cout << ans.size() << endl;
 return 0;
}



### Prompt

Develop a solution in Cpp to the problem described below:

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 A 3
#define AB 9

struct pazzle{
  int f[AB];
  int space;
  string path;

  bool operator < (const pazzle &p) const{
    for(int i=0;i<AB;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(pazzle p){
  for(int i=0;i<AB;i++)
    if(p.f[i]!=(i+1))return false;
  return true;
}

string bfs(pazzle s){
  queue<pazzle> Q;
  map<pazzle,bool> V;
  pazzle 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/A;
    int sy = u.space%A;

    for(int r=0;r<4;r++){
      int tx = sx+dx[r];
      int ty = sy+dy[r];

      if(tx<0 || ty<0 || tx>=A || ty>=A)continue;

      v=u;
      swap(v.f[u.space],v.f[tx*A+ty]);
      v.space = tx*A+ty;
      if(!V[v]){
        V[v]=true;
        v.path += dir[r];
        Q.push(v);
      }
    }
  }
  return "unsolvable";
}

int main(){
  pazzle in;

  for(int i=0;i<AB;i++){
    cin >> in.f[i];
    if(in.f[i]==0){
      in.f[i]=AB;
      in.space =i;
    }
  }
 string ans = bfs(in);
 cout << ans.size() << endl;
 return 0;
}



```