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

using namespace std;

#define n 4
#define n2 16
#define LIM 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[n2][n2];

struct puz {int f[n2], space, MD; };

puz state;
int limit;
int path[LIM];

int getAllMD(puz 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;
}

bool isSolved(){
  for(int i ; i<n2 ; 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/n;
  int sy=state.space%n;
  puz tmp;
  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;
    if(max(prev, r)-min(prev, r)==2)continue;
    tmp=state;
    state.MD-=MDT[tx*n+ty][state.f[tx*n+ty]-1];
    state.MD+=MDT[sx*n+sy][state.f[tx*n+ty]-1];
    swap(state.f[tx*n+ty], state.f[sx*n+sy]);
    state.space=tx*n+ty;
    if(dfs(depth+1,r)){path[depth]=r; return true; }
    state=tmp;
  }
  return false;
}

string indep(puz in){
  in.MD=getAllMD(in);
  for(limit=in.MD ; limit<=LIM ; limit++){
    state=in;
    if(dfs(0, -100)){
      string ans="";
      for(int i=0 ; i<limit ; i++)ans+=dir[path[i]];
      return ans;
    }
  }
  return "unsolvabe";
}

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);
  puz in;
  for(int i=0 ; i<n2 ; i++){
    cin>>in.f[i];
    if(in.f[i]==0){
      in.f[i]=n2;
      in.space=i;
    }
  }
  string ans=indep(in);
  cout<<ans.size()<<endl;
  return 0;
}

