#include <cstdio>           // printf(), scanf()
#include <string>
#include <queue>
#include <map>
#include <algorithm>        // swap()

using namespace std;
static const int N = 3;
static const int 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])
				return f[i] > p.f[i];
		}

		return false;
	}
};

bool
is_target(puzzle p)
{
	for (int i = 0; i < N2; ++i)
	{
		if (p.f[i] != i + 1)
			return false;
	}

	return true;
}

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

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 (is_target(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(int argc, char** argv)
{
	puzzle in;

	for (int i = 0; i < N2; ++i)
	{
		scanf("%d", &in.f[i]);
		if (in.f[i] == 0)
		{
			in.f[i] = N2;
			in.space = i;
		}
	}

	string ans = bfs(in);
	printf("%lu\n", ans.size());
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 <cstdio>           // printf(), scanf()
#include <string>
#include <queue>
#include <map>
#include <algorithm>        // swap()

using namespace std;
static const int N = 3;
static const int 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])
				return f[i] > p.f[i];
		}

		return false;
	}
};

bool
is_target(puzzle p)
{
	for (int i = 0; i < N2; ++i)
	{
		if (p.f[i] != i + 1)
			return false;
	}

	return true;
}

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

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 (is_target(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(int argc, char** argv)
{
	puzzle in;

	for (int i = 0; i < N2; ++i)
	{
		scanf("%d", &in.f[i]);
		if (in.f[i] == 0)
		{
			in.f[i] = N2;
			in.space = i;
		}
	}

	string ans = bfs(in);
	printf("%lu\n", ans.size());
	return 0;
}
```