#include <iostream>
#include <stdio.h>
#include <cmath>
using namespace std;
#define N 3 
#define PN 9
#define LIMIT 50

int MDT[PN][PN];
int dx[4] = {0,-1,0,1};
int dy[4] = {1,0,-1,0};
int limit;

struct Puzzle{ int f[PN], space, MD;};
Puzzle state;

int sumMD(Puzzle p){
    int sum = 0;
    for(int i = 0; i < PN; i++){
        if(p.f[i] == PN) continue;
        sum += MDT[i][p.f[i]-1];
    }
    return sum;
}

bool dfs(int depth, int prev){
    if(state.MD == 0) return true;
    if(depth+state.MD > limit) return false;

    int sx = state.space/N;
    int sy = state.space%N;
    Puzzle tmp = state;

    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;
        if((r+2)%4 == prev) continue;
        tmp = state;
        state.MD -= MDT[tx*N+ty][state.f[tx*N+ty]-1];
        state.MD += MDT[sx*N+sy][state.f[tx*N+ty]-1];
        swap(state.f[tx*N+ty],state.f[sx*N+sy]);
        state.space = tx*N+ty;
        if(dfs(depth+1,r)) return true;
        //if(dfs(depth+1,r)){path[depth] = r; return true;}
        state = tmp;
    }

    return false;
}

int main(){
    Puzzle p;

    for(int i = 0; i < PN; i++){
        cin >> p.f[i];
        if(p.f[i] == 0){
            p.f[i] = PN;
            p.space = i;
        }
    }

    for(int i = 0; i < PN; i++){
        for(int j = 0; j < PN; j++){
            MDT[i][j] = abs(i/N - j/N) + abs(i%N - j%N);
        }
    }

    p.MD = sumMD(p);

    for(limit = 0; limit < LIMIT; limit++){
        state = p;
        if(dfs(0,-1)){
            cout << limit << endl;
            return 0;
        }
    }

    return 0;
}

### Prompt

Create a solution in Cpp for 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 <stdio.h>
#include <cmath>
using namespace std;
#define N 3 
#define PN 9
#define LIMIT 50

int MDT[PN][PN];
int dx[4] = {0,-1,0,1};
int dy[4] = {1,0,-1,0};
int limit;

struct Puzzle{ int f[PN], space, MD;};
Puzzle state;

int sumMD(Puzzle p){
    int sum = 0;
    for(int i = 0; i < PN; i++){
        if(p.f[i] == PN) continue;
        sum += MDT[i][p.f[i]-1];
    }
    return sum;
}

bool dfs(int depth, int prev){
    if(state.MD == 0) return true;
    if(depth+state.MD > limit) return false;

    int sx = state.space/N;
    int sy = state.space%N;
    Puzzle tmp = state;

    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;
        if((r+2)%4 == prev) continue;
        tmp = state;
        state.MD -= MDT[tx*N+ty][state.f[tx*N+ty]-1];
        state.MD += MDT[sx*N+sy][state.f[tx*N+ty]-1];
        swap(state.f[tx*N+ty],state.f[sx*N+sy]);
        state.space = tx*N+ty;
        if(dfs(depth+1,r)) return true;
        //if(dfs(depth+1,r)){path[depth] = r; return true;}
        state = tmp;
    }

    return false;
}

int main(){
    Puzzle p;

    for(int i = 0; i < PN; i++){
        cin >> p.f[i];
        if(p.f[i] == 0){
            p.f[i] = PN;
            p.space = i;
        }
    }

    for(int i = 0; i < PN; i++){
        for(int j = 0; j < PN; j++){
            MDT[i][j] = abs(i/N - j/N) + abs(i%N - j%N);
        }
    }

    p.MD = sumMD(p);

    for(limit = 0; limit < LIMIT; limit++){
        state = p;
        if(dfs(0,-1)){
            cout << limit << endl;
            return 0;
        }
    }

    return 0;
}

```