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;
const int N=4;
const int N2=16;
const int LIMIT=100;
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[N2][N2];
struct Puzzle{int f[N2],space,MD;};
Puzzle state;
int limit;
int path[LIMIT];
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 isSolved()
{
	for(int i=0;i<N2;i++)
	    if(state.f[i]!=i+1)
	        return false;
	return true;
}
bool dfs(int depth,int prev)
{
	//cout<<depth<<' '<<prev<<endl;
	if(state.MD==0)
	    return true;
	if(depth+state.MD>limit)
	{
		//cout<<depth<<' '<<state.MD<<' '<<limit<<endl;
		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;
		//cout<<tx<<' '<<ty<<' '<<sx<<' '<<sy<<endl;
		if(max(prev,r)-min(prev,r)==2)
		    continue;
		tmp=state;
		//cout<<MDT[tx*N+ty][state.f[tx*N+ty]-1]<<' '
		//<<MDT[sx*N+sy][state.f[tx*N+ty]-1]<<endl;
		state.MD-=MDT[tx*N+ty][state.f[tx*N+ty]-1];
		state.MD+=MDT[sx*N+sy][state.f[tx*N+ty]-1];
		//cout<<state.MD<<endl;
		swap(state.f[tx*N+ty],state.f[sx*N+sy]);
		state.space=tx*N+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++)
	{
		//cout<<limit<<endl; 
		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<N2;i++)
	{
		for(int j=0;j<N2;j++)
		{
			MDT[i][j]=abs(i/N-j/N)+abs(i%N-j%N);
		    //cout<<MDT[i][j]<<' ';
	    }    
	    //cout<<endl;
	} 	    
	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;
		}
	}
	///cout<<in.space<<endl;
	string ans=iterative_deepening(in);
	cout<<ans.size()<<endl;
	return 0;
}