#include<iostream>
#include<map>
#include<queue>
#include<string>
using namespace std;

static const int N = 3;
static const int N2 = 9;

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'};

struct Puzzle {
	int p[N2], space;
	string path;

	bool operator < (const Puzzle &p1) const { //????????§?????????
		for (int i = 0; i < N2; i++) {
			if (p[i] == p1.p[i]) continue;
			return p[i] > p1.p[i];
		}
		return false;
	}
};

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

string bfs(Puzzle s) {
	queue<Puzzle> Q;
	map<Puzzle, bool> V; //????????´???Puzzle??§????????????????????§ < ????????????????????¨????????????

	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; //space??????
		int sy = u.space % N; //space??????
		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.p[u.space], v.p[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.p[i];
		if (!in.p[i]) {
			in.p[i] = N2; //isTarget?????????
			in.space = i;
		}
	}

	string ans = bfs(in);
	cout << ans.size() << endl;

	return 0;
}
### Prompt

Please formulate 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<map>
#include<queue>
#include<string>
using namespace std;

static const int N = 3;
static const int N2 = 9;

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'};

struct Puzzle {
	int p[N2], space;
	string path;

	bool operator < (const Puzzle &p1) const { //????????§?????????
		for (int i = 0; i < N2; i++) {
			if (p[i] == p1.p[i]) continue;
			return p[i] > p1.p[i];
		}
		return false;
	}
};

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

string bfs(Puzzle s) {
	queue<Puzzle> Q;
	map<Puzzle, bool> V; //????????´???Puzzle??§????????????????????§ < ????????????????????¨????????????

	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; //space??????
		int sy = u.space % N; //space??????
		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.p[u.space], v.p[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.p[i];
		if (!in.p[i]) {
			in.p[i] = N2; //isTarget?????????
			in.space = i;
		}
	}

	string ans = bfs(in);
	cout << ans.size() << endl;

	return 0;
}
```