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<cstdio>
#include<iostream>
#include<cmath>
#include<map>
#include<queue>
using namespace std;
#define N 4
#define N2 16


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;
  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);
    }
  }
  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;
}