#include<iostream>
#include<cmath>
#include<string>
#include<map>
#include<queue>
using namespace std;
#define N 3
#define 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])continue;
      return f[i] > p.f[i];
    }
    return false;
  }
};

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

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 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() {
  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

Construct a Cpp code solution to the problem outlined:

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<cmath>
#include<string>
#include<map>
#include<queue>
using namespace std;
#define N 3
#define 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])continue;
      return f[i] > p.f[i];
    }
    return false;
  }
};

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

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 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() {
  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;
}
```