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]; //Total manhattan distance

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;
  }
};

//各パネルにおける最終状態までのマンハッタン距離の総和を返す
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(){
  //始点iから終点jまでのマンハッタン距離を全ての組み合わせで計算
  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;
}