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<bits/stdc++.h>
using namespace std;
#define N 4
#define N2 16
#define LIM 100
int dx[4]={0,-1,0,1};
int dy[4]={1,0,-1,0};
char dir[4]={'r','u','l','d'};
int MDT[N2][N2];
struct puzzle{int f[N2],space,MD;};
puzzle state;
int lim;
int path[LIM];
int getallMD(puzzle pz){
  int sum=0;
  for(int i=0;i<N2;i++){
    if(pz.f[i]==N2)continue;
    sum+=MDT[i][pz.f[i]-1];
  }
  return sum;
}
bool solved(){
  for(int i=0;i<N2;i++)if(state.f[i]!=i+1)return false;
  return true;
}
bool dfs(int de,int pr){
  if(state.MD==0)return true;
  if(de+state.MD>lim)return false;
  int sx=state.space/N;
  int sy=state.space%N;
  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>=N||ty>=N)continue;
    if(max(pr,r)-min(pr,r)==2)continue;
    tmp=state;
    state.MD-=MDT[tx*N+ty][state.f[tx*N+ty]-1];
    state.MD+=MDT[sx*N+sy][state.f[tx*N+ty]-1];
    swap(state.f[tx*N+ty],state.f[sx*N+sy]);
    state.space=tx*N+ty;
    if(dfs(de+1,r)){
      path[de]=r;
      return true;
    }
    state=tmp;
  }
  return false;
}
string iterative(puzzle in){
  in.MD = getallMD(in);
  for(lim=in.MD;lim<=LIM;lim++){
    state=in;
    if(dfs(0,-100)){
      string ans="";
      for(int i=0;i<lim;i++)ans+=dir[path[i]];
      return ans;
    }
  }
  return "unsolvable";
}
int main(){
  for(int i=0;i<N2;i++)
    for(int j=0;j<N2;j++)
      MDT[i][j]=abs(i/N-j/N)+abs(i%N-j%N);
  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 = iterative(in);
  cout << ans.size() << endl;
  return 0;
}