#include <bits/stdc++.h>
using namespace std;

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

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]) continue;
            return f[i] < p.f[i];
        }
        return false;
    }
};

bool isTarget(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 (isTarget(u)) return u.path;

        int sx = u.space % N;
        int sy = u.space / N;
        for (int r = 0; r < 4; r++) {
            int nx = sx + dx[r];
            int ny = sy + dy[r];
            if (nx < 0 || ny < 0 || nx >= N || ny >= N) continue;

            v = u;
            swap(v.f[u.space], v.f[ny*N + nx]);
            v.space = ny*N + nx;
            if (!V[v]) {
                V[v] = true;
                v.path += dir[r];
                Q.push(v);
            }
        }
    }

    return "unsolvable";
}

int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    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

Develop a solution in Cpp 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 <bits/stdc++.h>
using namespace std;

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

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]) continue;
            return f[i] < p.f[i];
        }
        return false;
    }
};

bool isTarget(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 (isTarget(u)) return u.path;

        int sx = u.space % N;
        int sy = u.space / N;
        for (int r = 0; r < 4; r++) {
            int nx = sx + dx[r];
            int ny = sy + dy[r];
            if (nx < 0 || ny < 0 || nx >= N || ny >= N) continue;

            v = u;
            swap(v.f[u.space], v.f[ny*N + nx]);
            v.space = ny*N + nx;
            if (!V[v]) {
                V[v] = true;
                v.path += dir[r];
                Q.push(v);
            }
        }
    }

    return "unsolvable";
}

int main(void) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    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;
}
```