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

struct puzzle{
  int x[N2];
  int space;
  string path;
  bool operator < (const puzzle &p)const{
    for(int i=0;i<N2;i++){
      if(x[i]==p.x[i]) continue;
      return x[i]>p.x[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 target(puzzle);
string bfs(puzzle);

int main(){
  int i;
  string answer;
  puzzle in;

  for(i=0;i<N2;i++){
    cin >> in.x[i];
    if(in.x[i] == 0){
      in.x[i] = N2;
      in.space = i;
    }
  }
  answer = bfs(in);
  cout << answer.size() << endl;

  return 0;
}

bool target(puzzle p){
  int i;
  for(i=0;i<N2;i++)
    if(p.x[i] != (i+1)) return false;
  return true;
}

string bfs(puzzle p){
  int px,py,tx,ty,r;
  queue<puzzle> Que;
  map<puzzle,bool> V;
  puzzle m,k;
  p.path="";
  Que.push(p);
  V[p]=true;

  while(!Que.empty()){
    m=Que.front(); Que.pop();
    if(target(m)==true) return m.path;
    px=m.space/N;
    py=m.space%N;
    for(r=0;r<4;r++){
      tx=px+dx[r];
      ty=py+dy[r];
      if(tx<0 || ty<0 || tx>=N || ty>=N) continue;
      k=m;
      swap(k.x[m.space],k.x[tx*N+ty]);
      k.space=tx*N+ty;
      if(!V[k]){
	V[k]=true;
	k.path += dir[r];
	Que.push(k);
      }
    }
  }
  return "unsolvable";
}


### Prompt

In cpp, your task is to solve 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 x[N2];
  int space;
  string path;
  bool operator < (const puzzle &p)const{
    for(int i=0;i<N2;i++){
      if(x[i]==p.x[i]) continue;
      return x[i]>p.x[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 target(puzzle);
string bfs(puzzle);

int main(){
  int i;
  string answer;
  puzzle in;

  for(i=0;i<N2;i++){
    cin >> in.x[i];
    if(in.x[i] == 0){
      in.x[i] = N2;
      in.space = i;
    }
  }
  answer = bfs(in);
  cout << answer.size() << endl;

  return 0;
}

bool target(puzzle p){
  int i;
  for(i=0;i<N2;i++)
    if(p.x[i] != (i+1)) return false;
  return true;
}

string bfs(puzzle p){
  int px,py,tx,ty,r;
  queue<puzzle> Que;
  map<puzzle,bool> V;
  puzzle m,k;
  p.path="";
  Que.push(p);
  V[p]=true;

  while(!Que.empty()){
    m=Que.front(); Que.pop();
    if(target(m)==true) return m.path;
    px=m.space/N;
    py=m.space%N;
    for(r=0;r<4;r++){
      tx=px+dx[r];
      ty=py+dy[r];
      if(tx<0 || ty<0 || tx>=N || ty>=N) continue;
      k=m;
      swap(k.x[m.space],k.x[tx*N+ty]);
      k.space=tx*N+ty;
      if(!V[k]){
	V[k]=true;
	k.path += dir[r];
	Que.push(k);
      }
    }
  }
  return "unsolvable";
}


```