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 <stdio.h>
#include <iostream>
#include <cmath>
#include <string>
#include <cassert>

using namespace std;

#define MAX 4
#define MAX2 16
#define LIMIT 100

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[MAX2][MAX2];
struct Puzzle{int f[MAX2],space,MD;};

Puzzle state;
int limit;
int path[LIMIT];

int getALLMD(Puzzle p){
  int sum=0;
  for(int i=0;i<MAX2;i++){
    if(p.f[i]==MAX2)continue;
    sum += MDT[i][p.f[i]-1];
  }
  return sum;
}
bool isSolved(){
  for(int i=0;i<MAX2;i++)if(state.f[i]!=i+1)return false;
  return true;
}
bool DFS(int depth,int prev){
  if(state.MD==0)return true;
  if(depth+state.MD>limit)return false;

  int sx=state.space/MAX;
  int sy=state.space%MAX;
  Puzzle tmp;

  for(int r=0;r<4;r++){
    int tx=sx+dx[r];
    int ty=sy+dy[r];
    if(tx<0 || ty<0 || tx>=MAX || ty>=MAX)continue;
    if(max(prev,r)-min(prev,r)==2)continue;
    tmp=state;
    state.MD -= MDT[tx*MAX+ty][state.f[tx*MAX+ty]-1];
    state.MD += MDT[sx*MAX+sy][state.f[tx*MAX+ty]-1];
    swap(state.f[tx*MAX+ty],state.f[sx*MAX+sy]);
    state.space=tx*MAX+ty;
    if(DFS(depth+1,r)){
      path[depth]=r;
      return true;
    }
    state=tmp;
  }
  return false;
}

string iterative_deepening(Puzzle in){
  in.MD=getALLMD(in);
  for(limit=in.MD;limit<=LIMIT;limit++){
    state=in;
    if(DFS(0,-100)){
      string ans = "";
      for(int i=0;i<limit;i++)ans+=dir[path[i]];
      return ans;
    }
  }
  return "unsolvable";
}

int main() {
  for(int i=0;i<MAX2;i++)
    for(int j=0;j<MAX2;j++)
      MDT[i][j]=abs(i/MAX-j/MAX)+abs(i%MAX-j%MAX);
      
  Puzzle in;

  for(int i=0;i<MAX2;i++){
    cin >> in.f[i];
    if(in.f[i]==0){
      in.f[i]=MAX2;
      in.space=i;
    }
  }
  string ans = iterative_deepening(in);
  cout << ans.size() << endl;
  
  return 0;
}