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<cassert>
#include<string>

using namespace std;
#define N 4
#define N2 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[N2][N2];


struct Puzzle{int f[N2],space,MD;};

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

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;
}



bool isSolved(){
  for(int i = 0; i<N2;i++)
    if(state.f[i] != i+1) return false;
  return true;
}


bool isTarget(Puzzle p){
  for(int i = 0; i<N2;i++){
    if(p.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;
  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>=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 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<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;
    }
  }
  string ans = iterative_deepening(in);
  cout<<ans.size()<<endl;

  return 0;
}