Problem: The goal of the 15 puzzle problem is to complete pieces on $4 \times 4$ 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 15 as shown below.


1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15


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
9 10 11 12
13 14 15 0


Write a program which reads an initial state of the puzzle and prints the fewest steps to solve the puzzle.

Constraints

* The given puzzle is solvable in at most 45 steps.

Input

The $4 \times 4$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Example

Input

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15


Output

8

Solution (in CPP):

Solution:
#include<iostream>
#include<cmath>
#include<string>
#include<queue>
#include<map>
#define LIMIT 100

using namespace std;

static const int dx[4] = {0,-1,0,1};
static const int dy[4] = {1,0,-1,0};
static const char dir[4] = {'r','u','l','d'};
int mdt[16][16];

class puzzle{
public :
  int f[16],space,md;
};

puzzle state;
int limit;
int path[LIMIT];

int getAllMD(puzzle pz){
  int sum =0;
  for(int i=0;i<16;i++){
    if(pz.f[i]==16)continue;
    sum+=mdt[i][pz.f[i]-1];
  }
  return sum;
}

bool isSolved(){
  for(int i=0;i<16;i++)if(state.f[i]!=i+1) return false;
  return true;
}

bool dfs(int depth,int prev){
  if(state.md==0)return true;
  if(depth+state.md >limit) return false;

  int sx=state.space/4;
  int sy=state.space%4;
  puzzle tmp;

  for(int r=0;r<4;r++){
    int tx=sx+dx[r];
    int ty=sy+dy[r];
    if(tx<0||ty<0||tx>=4||ty>=4)continue;
    if(max(prev,r)-min(prev,r)==2)continue;
    tmp=state;

    state.md-=mdt[tx*4+ty][state.f[tx*4+ty]-1];
    state.md+=mdt[sx*4+sy][state.f[tx*4+ty]-1];
    swap(state.f[tx*4+ty],state.f[sx*4+sy]);
    state.space = tx*4+ty;
    if(dfs(depth+1,r)){path[depth]=r;return true;}
    state = tmp;
  }
  return false;
}
string iterative_deepening(puzzle in){
  in.md = getAllMD(in);

  for(limit = in.md;limit<=LIMIT;limit++){
    state = in;
    if(dfs(0,-100)){
      string ans = "";
      for(int i=0;i<limit;i++)ans +=dir[path[i]];
      return ans;
    }
  }
  return "unsolvable";
}

int main(){
  for(int i=0;i<16;i++){
    for(int j=0;j<16;j++){
      mdt[i][j]=abs(i/4-j/4)+abs(i%4-j%4);
    }
  }
  puzzle in;
  for(int i=0;i<16;i++){
    cin >> in.f[i];
    if(in.f[i]==0){
      in.f[i]=16;
      in.space =i;
    }
  }
  string ans =iterative_deepening(in);
  cout << ans.size()<<endl;

  return 0;
}