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

struct Puzzle {
    int f[9];
    int space;
    string path = "";

    bool operator < (const Puzzle& p) const {
        for(int i = 0; i < 9; i++) {
            if(f[i] == p.f[i]) continue;
            return f[i] > p.f[i];
        }
        return false;
    }
};

const int nxt[][2] = {{-1,0},{0,-1},{1,0},{0,1} };
const char dir[] = {'u','l','d','r'};

inline bool isTarget(Puzzle p) {
    for(int i = 0; i < 9; i++) {
        if(p.f[i] != (i+1)) return false;
    }
    return true;
}

string bfs(Puzzle s) {
    queue<Puzzle> Q;
    map<Puzzle,bool> V; //K-V对表示哪个迷宫(状态)是已经搜索过的了
    Q.push(s);
    V[s] = true;

    while(!Q.empty()) {
        Puzzle u = Q.front(); Q.pop();
        if(isTarget(u)) return u.path;
        int sx = u.space / 3;
        int sy = u.space % 3; // 得到空位所在的行列
        for(int i = 0; i < 4; i++) {
            int tx = sx + nxt[i][0];
            int ty = sy + nxt[i][1]; // 空位旁边所有拼图片
            if(tx < 0 || ty < 0 || tx >= 3 || ty >= 3) continue;
            Puzzle v = u;
            swap(v.f[u.space],v.f[tx*3 + ty]);
            v.space = tx * 3 + ty;
            if(!V[v]) {
                V[v] = true;
                v.path += dir[i];
                Q.push(v);
            }
        }
    }
}

int main() {
    Puzzle in;
    for(int i =0; i < 9; i++) {
        cin >> in.f[i];
        if(in.f[i] == 0) {
            in.f[i] = 9;
            in.space = i;
        }
    }
    string ans = bfs(in);
    cout << ans.size() << endl;
    return 0;
}


### Prompt

Create a solution in cpp for 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 <bits/stdc++.h>
using namespace std;

struct Puzzle {
    int f[9];
    int space;
    string path = "";

    bool operator < (const Puzzle& p) const {
        for(int i = 0; i < 9; i++) {
            if(f[i] == p.f[i]) continue;
            return f[i] > p.f[i];
        }
        return false;
    }
};

const int nxt[][2] = {{-1,0},{0,-1},{1,0},{0,1} };
const char dir[] = {'u','l','d','r'};

inline bool isTarget(Puzzle p) {
    for(int i = 0; i < 9; i++) {
        if(p.f[i] != (i+1)) return false;
    }
    return true;
}

string bfs(Puzzle s) {
    queue<Puzzle> Q;
    map<Puzzle,bool> V; //K-V对表示哪个迷宫(状态)是已经搜索过的了
    Q.push(s);
    V[s] = true;

    while(!Q.empty()) {
        Puzzle u = Q.front(); Q.pop();
        if(isTarget(u)) return u.path;
        int sx = u.space / 3;
        int sy = u.space % 3; // 得到空位所在的行列
        for(int i = 0; i < 4; i++) {
            int tx = sx + nxt[i][0];
            int ty = sy + nxt[i][1]; // 空位旁边所有拼图片
            if(tx < 0 || ty < 0 || tx >= 3 || ty >= 3) continue;
            Puzzle v = u;
            swap(v.f[u.space],v.f[tx*3 + ty]);
            v.space = tx * 3 + ty;
            if(!V[v]) {
                V[v] = true;
                v.path += dir[i];
                Q.push(v);
            }
        }
    }
}

int main() {
    Puzzle in;
    for(int i =0; i < 9; i++) {
        cin >> in.f[i];
        if(in.f[i] == 0) {
            in.f[i] = 9;
            in.space = i;
        }
    }
    string ans = bfs(in);
    cout << ans.size() << endl;
    return 0;
}


```