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>

#define N 4
#define N2 16
#define LIMIT 100

using namespace std;

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<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;

}

