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
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<cassert>
using namespace std;
const int dx[4]={0,-1,0,1};
const int dy[4]={1,0,-1,0};
const char dir[4]={'r','u','l','d'};
const int N=4;
const int N2=16;
const int LIMIT=45;
int MDT[N2][N2];
struct Puzzle
{	
	int f[N2],space;
	int 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);
		else sum+=MDT[i][pz.f[i]-1];
	}
	
	return sum;
}
int check(int x,int y)
{
	if(x<0||y<0||x>=N||y>=N)return 0;
	return 1;
}
int dfs(int depth,int prev)
{
	if(state.MD==0)return 1;
	if(depth+state.MD>limit)return 0;
	
	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(!check(tx,ty))continue;
		if(max(prev,r)-min(prev,r)==2)continue;
		
		tmp=state;
		
		int nt,ns;
		nt=tx*N+ty;
		ns=sx*N+sy;
		state.MD-=MDT[nt][state.f[nt]-1];
		state.MD+=MDT[ns][state.f[nt]-1];
		
		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 1;
		}
		state=tmp;
	}
	return 0;
}
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;
		}
    }
}
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_deepening(in);
	
	cout<<ans.size()<<endl;
	return 0;
}