#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 "usdolvable";
}

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

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 "usdolvable";
}

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;
}
```