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 <iostream>
#include <stdio.h>
#include <cmath>
#include <string>
#include <cassert>
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 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<N; 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;
}

