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>
using namespace std;
#define N1 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 / N1;
  int sy = state.space % N1;
  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 >= N1 || ty >= N1) continue;
    if(max(prev,r)-min(prev,r) == 2) continue;
    tmp = state;
    state.MD -= MDT[tx*N1 + ty][state.f[tx*N1 + ty] - 1];
    state.MD += MDT[sx*N1 + sy][state.f[tx*N1 + ty] - 1];
    swap(state.f[tx*N1 +ty],state.f[sx*N1 + sy]);
    state.space = tx*N1 + 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/N1 - j/N1) + abs(i%N1 - j%N1);
    }
  }

  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;
}

