#include<iostream>   
#include<cstdio>  
#include<algorithm>
#include<queue>
#include<map>
#include<string>
using namespace std;
const int row = 3;
const int maxx = 9;

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

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

bool isTarget(Puzzle p) {
	for (int i = 0; i < maxx; 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 (isTarget(u))
			return u.path;
		int sx = u.space / row;
		int sy = u.space % row;
		for (int r = 0; r < 4; r++) {
			int tx = sx + dx[r];
			int ty = sy + dy[r];

			if (tx < 0 || ty < 0 || tx >= row || ty >= row)
				continue;
			v = u;
			swap(v.f[u.space], v.f[tx * row + ty]);
			v.space = tx * row + ty;
			if (!V[v]) {
				V[v] = true;
				v.path += dir[r];
				Q.push(v);
			}
		}
	}
	return "unsolveable";
}
int main()
{
	Puzzle in;
	for (int i = 0; i < maxx; i++) {
		cin >> in.f[i];
		if (in.f[i] == 0) {
			in.f[i] = maxx;
			in.space = i;
		}
	}
	string ans = bfs(in);
	printf("%d\n", ans.size());
	return 0;
}

### Prompt

In CPP, your task is to solve 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<cstdio>  
#include<algorithm>
#include<queue>
#include<map>
#include<string>
using namespace std;
const int row = 3;
const int maxx = 9;

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

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

bool isTarget(Puzzle p) {
	for (int i = 0; i < maxx; 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 (isTarget(u))
			return u.path;
		int sx = u.space / row;
		int sy = u.space % row;
		for (int r = 0; r < 4; r++) {
			int tx = sx + dx[r];
			int ty = sy + dy[r];

			if (tx < 0 || ty < 0 || tx >= row || ty >= row)
				continue;
			v = u;
			swap(v.f[u.space], v.f[tx * row + ty]);
			v.space = tx * row + ty;
			if (!V[v]) {
				V[v] = true;
				v.path += dir[r];
				Q.push(v);
			}
		}
	}
	return "unsolveable";
}
int main()
{
	Puzzle in;
	for (int i = 0; i < maxx; i++) {
		cin >> in.f[i];
		if (in.f[i] == 0) {
			in.f[i] = maxx;
			in.space = i;
		}
	}
	string ans = bfs(in);
	printf("%d\n", ans.size());
	return 0;
}

```