#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 "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 provide a cpp coded solution 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 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 "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;
}


```