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

using namespace std;

#define N 3
#define N2 9

struct Puzzle {
	int f[N2], space;
	string path;
	bool operator<(const Puzzle& a) const {
		for (int i = 0; i < N2; ++i) {
			if (f[i] == a.f[i]) continue;
			return f[i] > a.f[i];
		}
		return false;
	}
};

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

bool ok(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 (ok(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 || 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;
			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 <string>
#include <queue>
#include <map>

using namespace std;

#define N 3
#define N2 9

struct Puzzle {
	int f[N2], space;
	string path;
	bool operator<(const Puzzle& a) const {
		for (int i = 0; i < N2; ++i) {
			if (f[i] == a.f[i]) continue;
			return f[i] > a.f[i];
		}
		return false;
	}
};

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

bool ok(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 (ok(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 || 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;
			in.space = i;
		}
	}
	string ans = bfs(in);
	cout << ans.size() << endl;

	return 0;
}

```