#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 int 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; // set space
			in.space = i;
		}
	}
	string ans = bfs(in);
	cout << ans.size() <<  endl;
	
	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 <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 int 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; // set space
			in.space = i;
		}
	}
	string ans = bfs(in);
	cout << ans.size() <<  endl;
	
	return 0;
}


```