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 NN 16
#define LIMIT 45

struct P{
  int f[NN];
  int t;
  int st;
  bool operator < (const P &p) const{
    for(int i=0;i<NN;i++){
      if(f[i]==p.f[i])continue;
      return f[i]>p.f[i];
    }
    return false;
  }
};

struct Cost{
  P p;
  int cost;
  bool operator < (const Cost &c) const{
    return cost+p.t > c.cost+c.p.t;
  }
};

int man(int f[]){
  int ans=0;
  for(int i=0;i<NN;i++){
    if(f[i]==0)continue;
    ans+=abs((f[i]-1)/N-i/N);
    ans+=abs((f[i]-1)%N-i%N);
  }
  return ans;
}

int main(){
  P in;
  map<P,bool> V;
  priority_queue<Cost> Q;
  for(int i=0;i<NN;i++){
    cin>>in.f[i];
    if(in.f[i]==0)in.st=i;
  }
  in.t=0;
  Cost t;
  t.p=in;
  t.cost=man(t.p.f);
  Q.push(t);
  while(!Q.empty()){
    P top=Q.top().p;
    Q.pop();

    if(man(top.f)+top.t<=LIMIT){
      V[top]=true;
      int flag=0;
      for(int i=0;i<NN;i++){
	if((i+1)%NN==top.f[i])flag++;
      }
      if(flag==NN){
	cout<<top.t<<endl;
	break;
      }
      int st=top.st;
      top.t++;
      P tmp;
      if(st-N>=0){
	tmp=top;
	swap(tmp.f[st],tmp.f[st-N]);
	tmp.st-=N;
	if(!V[tmp]&&man(tmp.f)+tmp.t<=LIMIT){
	  V[tmp]=true;
	  Cost t;
	  t.p=tmp;
	  t.cost=man(t.p.f);
	  Q.push(t);
	}
      }
      if(st+N<NN){
	tmp=top;
	swap(tmp.f[st],tmp.f[st+N]);
	tmp.st+=N;
	if(!V[tmp]&&man(tmp.f)+tmp.t<=LIMIT){
	  V[tmp]=true;
	   Cost t;
	  t.p=tmp;
	  t.cost=man(t.p.f);
	  Q.push(t);
	}
      }
      if(st%N!=N-1){
	tmp=top;
	swap(tmp.f[st],tmp.f[st+1]);
	tmp.st++;
	if(!V[tmp]&&man(tmp.f)+tmp.t<=LIMIT){
	  V[tmp]=true;
	  Cost t;
	  t.p=tmp;
	  t.cost=man(t.p.f);
	  Q.push(t);
	}
      }
      if(st%N!=0){
	tmp=top;
	swap(tmp.f[st],tmp.f[st-1]);
	tmp.st--;
	if(!V[tmp]&&man(tmp.f)+tmp.t<=LIMIT){
	  V[tmp]=true;
	  Cost t;
	  t.p=tmp;
	  t.cost=man(t.p.f);
	  Q.push(t);
	}
      }
    }
  }
  return 0;
}