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<cstdio>
#include<cmath>
#include<string>
#include<iostream>
#define N 4
#define N2 16
#define LIMIT 100
using namespace std;
static const char c[4] = {'r','u', 'l', 'd'};
static const int x[4] = {0,-1, 0, 1};
static const int y[4] = {1,0, -1, 0};
int limit, MDT[N2][N2], path[LIMIT];
struct Puzzle
{
  int f[N2], space, MD;
};
Puzzle state;
int getMD(Puzzle p)
{
  int sum = 0;
  for(int i=0; i<N2; i++)
  {
    if(p.f[i] == N2) continue;
    sum += MDT[i][p.f[i] - 1];
  }
  return sum;
}

bool dfs(int depth, int previous)
{
  if(state.MD == 0) return true;
  if(depth + state.MD > limit) return false;
  Puzzle temp;
  int sx, sy, tx, ty;
  sx = state.space/N; sy = state.space%N;
  for(int r=0; r<4; r++)
  {
    tx = sx + x[r]; ty = sy + y[r];
    if(ty < 0 || tx < 0 || tx >=N || ty >= N) continue;
    if(max(previous, r) - min(previous, r) == 2) continue;
    temp = state;
    state.MD -= MDT[tx*N + ty][state.f[tx*N + ty]-1];
    state.MD += MDT[state.space][state.f[tx*N + ty] -1];
    swap(state.f[state.space], state.f[tx*N + ty]);
    state.space = tx*N + ty;
    if(dfs(depth+1, r))           //tai
    {
      path[depth] = r; return true;// xiu le
    }
    state = temp;
  }
  return false;
}
string iterative_deepening(Puzzle &p)
{
  p.MD = getMD(p);
  for(limit = p.MD; limit<=LIMIT; limit++)
  {
    state = p;
    if(dfs(0, -100))
    {
      string ans = "";
      for(int i=0; i<limit; i++)
      {
        ans += c[path[i]];
      }
      return ans;
    }
  }
  return "unsolvable";
}
int main()
{
  Puzzle in;
  for(int i=0; i<N2; i++)
  {
    scanf("%d", &in.f[i]);
    if(in.f[i] == 0)
    {
      in.f[i] = N2; in.space = i;
    }
  }
  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);
  }
  string ans = iterative_deepening(in);
  cout << ans.size() << endl;

  return 0;
}

