#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;
    int i;

    bool operator < (const Puzzle &p) const
    {
        int i;
        for(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)
{
    int i;

    for(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

Your task is to create a CPP solution 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>
using namespace std;

#define N 3
#define N2 9

struct Puzzle
{
    int f[N2];
    int space;
    string path;
    int i;

    bool operator < (const Puzzle &p) const
    {
        int i;
        for(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)
{
    int i;

    for(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;

}

```