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 LIMIT 100

const int dx[4]={0,-1,0,1};
const int dy[4]={1,0,-1,0};
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 dfs(int depth,int prev)
{
    if(state.MD==0){
        return true;
    }
    if(depth+state.MD>limit){
        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(prev,r)-min(prev,r)==2){
            continue;
        }
        tmp=state;
        //计算曼哈顿距离的差值，同时交换拼图块
        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(depth+1,r)){
            path[depth]=r;
            return true;
        }
        state=tmp;
    }
    return false;
}

string iterative_deepening(puzzle in)
{
    in.MD=getALLMD(in);//初始状态的曼哈顿距离
    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<N2;i++){
        for(int j=0;j<N2;j++){
            //曼哈顿距离，此法太妙
            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);
//    for(int i=0;i<limit;i++){
//        cout<<path[i]<<" ";
//    }
    cout<<ans.size()<<endl;
    return 0;
}