// 8 puzzle
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
#define N  3
#define N2 9
using namespace std;
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.space = i;
                in.f[i] = 9;
        }
    }
    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
// 8 puzzle
#include <iostream>
#include <algorithm>
#include <queue>
#include <map>
#define N  3
#define N2 9
using namespace std;
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.space = i;
                in.f[i] = 9;
        }
    }
    string ans = bfs(in);
    cout<<ans.size()<<endl;
    return 0;
}

```