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
//13_C 15パズル問題   15 Puzzle

#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 MTD[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 += MTD[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 -= MTD[tx*N+ty][v.f[tx*N+ty]-1];
      v.MD += MTD[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++)
      MTD[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;
}

