#include <iostream>
#include <map>
#include <queue>

using namespace std;
const int N = 3;
const int N2 = 9;
const int dx[4] = {-1,0,1,0};
const int dy[4] = {0,-1,0,1};
const char dir[4] = {'u','l','d','r'};


typedef struct Puzzle {
	int M[N2]; //状态 
	int space; //空格位置 
	string path; //记录空格的移动路径 
	
	bool operator<(const Puzzle & p) const {
		for (int i=0;i<N2;i++) {
			if (M[i] == p.M[i]) continue;
			return M[i] > p.M[i]; //为了让后面生成的状态靠前 
		}
		return false;
	}
	
} Puzzle;

bool isTarget(Puzzle & p) {
	for (int i=0;i<N2-1;i++) {
		if (p.M[i] != i+1) return false;
	}
	return true;
}

string bfs(Puzzle p) { //广度优先搜索 
	queue<Puzzle> Q;
	map<Puzzle,bool> V;
	Puzzle u,v;
	p.path = "";
	Q.push(p);
	V[p] = 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;
		//搜索四个方向 
		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; //不存在的情况
			int tspace = tx * N + ty; //新的空格位置
			v = u;//拷贝一份
			v.space = tspace; //改变数据
			swap(v.M[u.space],v.M[v.space]);
			//检查新生成的状态是否已经存在 
			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.M[i];
		if (in.M[i] == 0) {
			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 <iostream>
#include <map>
#include <queue>

using namespace std;
const int N = 3;
const int N2 = 9;
const int dx[4] = {-1,0,1,0};
const int dy[4] = {0,-1,0,1};
const char dir[4] = {'u','l','d','r'};


typedef struct Puzzle {
	int M[N2]; //状态 
	int space; //空格位置 
	string path; //记录空格的移动路径 
	
	bool operator<(const Puzzle & p) const {
		for (int i=0;i<N2;i++) {
			if (M[i] == p.M[i]) continue;
			return M[i] > p.M[i]; //为了让后面生成的状态靠前 
		}
		return false;
	}
	
} Puzzle;

bool isTarget(Puzzle & p) {
	for (int i=0;i<N2-1;i++) {
		if (p.M[i] != i+1) return false;
	}
	return true;
}

string bfs(Puzzle p) { //广度优先搜索 
	queue<Puzzle> Q;
	map<Puzzle,bool> V;
	Puzzle u,v;
	p.path = "";
	Q.push(p);
	V[p] = 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;
		//搜索四个方向 
		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; //不存在的情况
			int tspace = tx * N + ty; //新的空格位置
			v = u;//拷贝一份
			v.space = tspace; //改变数据
			swap(v.M[u.space],v.M[v.space]);
			//检查新生成的状态是否已经存在 
			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.M[i];
		if (in.M[i] == 0) {
			in.space = i;
		}
	}
	string ans = bfs(in);
	cout << ans.size() << endl; 
 	return 0;
}

```