#include<iostream>
#include<cmath>
#include<string>
#include<map>
#include<queue>
#include<cassert>
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);
      }
    }
  }
  assert(false);
  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

Please create a solution in CPP to 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>
#include<cassert>
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);
      }
    }
  }
  assert(false);
  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;
}

```