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

#define N 4
#define N2 16
#define LIMIT 100
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 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 ku(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 " ";
}
 
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 = ku(in);
  cout << ans.size() << endl;
 
  return 0;
}