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:
//A*
//优先对“起点到当前位置的成本+当前位置到目标状态的推测值”最小的状态进行状态迁移
#include<bits/stdc++.h>
 
using namespace std;
 
#define N 4
#define N2 16
 
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;
    int cost;
    bool operator < (const Puzzle &p)const{
        for(int i=0;i<N2;i++){
            if(f[i]==p.f[i]){
                continue;
            }
            return f[i]<p.f[i];
        }
        return false;
    }
};
 
struct state{
    Puzzle puzzle;
    int estimated;
    bool operator < (const state &s)const{
        return estimated>s.estimated;
    }
};
 
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;
}
 
int astar(Puzzle s)
{
    priority_queue<state>pq;
    s.MD=getALLMD(s);
    s.cost=0;
    map<Puzzle,bool>V;
    Puzzle u,v;
    state initial;
    initial.puzzle=s;
    initial.estimated=getALLMD(s);
    pq.push(initial);
    while(!pq.empty()){
        state st=pq.top();
        pq.pop();
        u=st.puzzle;
        if(u.MD==0){
            return u.cost;
        }
        V[u]=true;
        int sx=u.space/N;
        int sy=u.space%N;
        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;
            }
            v=u;
            v.MD-=MDT[tx*N+ty][v.f[tx*N+ty]-1];
            v.MD+=MDT[sx*N+sy][v.f[tx*N+ty]-1];
            swap(v.f[sx*N+sy],v.f[tx*N+ty]);
            v.space=tx*N+ty;
            if(!V[v]){
                v.cost++;
                state news;
                news.puzzle=v;
                news.estimated=v.cost+v.MD;
                pq.push(news);
            }
        }
    }
    return -1;
}
 
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);
        }
    }
//    for(int i=0;i<N2;i++){
//        for(int j=0;j<N2;j++){
//            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<<astar(in)<<endl;
    return 0;
}