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
#include<bits/stdc++.h>
using namespace std;
#define N 4
#define N2 16

static const int dx[4]={-1,0,1,0};
static const int dy[4]={0,-1,0,1};
//static const char dir[4]={'u','l','d','r'};
int MDT[N2][N2];

struct Puzzle{
  int f[N2],MD;
  int space,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;
}
priority_queue<state> PQ;
map<Puzzle,bool> V;
int astar(Puzzle &s){
 
  s.MD=getAllMD(s);
  s.cost=0;
 
  Puzzle v;
  state initial;
  initial.puzzle=s;
  initial.estimated=getAllMD(s);
  PQ.push(initial);

  while(!PQ.empty()){
    state st=PQ.top();PQ.pop();
    if(st.puzzle.MD==0) return st.puzzle.cost;
    int sx=st.puzzle.space/N;
    int sy=st.puzzle.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=st.puzzle;
      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[st.puzzle]=true;
	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;
}

