#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 3
#define N2 9
class puzzle{
public:
	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]) return f[i]>p.f[i];
		}
		return false;
	}
	
};

static const int dx[4]={-1,+1,0,0};
static const int dy[4]={0,0,-1,+1};
static const char dir[4]={'u','d','l','r'};

bool isTarget(puzzle u){
	for(int i=0;i<N2;i++){
		if(u.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,sy=u.space%N;
		for(int r=0;r<4;r++){
			int tx=sx+dx[r],ty=sy+dy[r];
			if(tx<0 || tx>=N || ty<0 || ty>=N) continue;
			v=u;
			swap(v.f[v.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

Construct a CPP code solution to the problem outlined:

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 <set>
#include <cmath>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define N 3
#define N2 9
class puzzle{
public:
	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]) return f[i]>p.f[i];
		}
		return false;
	}
	
};

static const int dx[4]={-1,+1,0,0};
static const int dy[4]={0,0,-1,+1};
static const char dir[4]={'u','d','l','r'};

bool isTarget(puzzle u){
	for(int i=0;i<N2;i++){
		if(u.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,sy=u.space%N;
		for(int r=0;r<4;r++){
			int tx=sx+dx[r],ty=sy+dy[r];
			if(tx<0 || tx>=N || ty<0 || ty>=N) continue;
			v=u;
			swap(v.f[v.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;
}

```