#include <iostream>
#include <cmath>
#include <string>
#include <map>
#include <queue>

using namespace std;
const int N = 3;
const int 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 or ty < 0 or tx >= N or 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;
const int N = 3;
const int 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 or ty < 0 or tx >= N or 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 ;
}
```