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<cstdio>
#include<iostream>
#include<cmath>
#include<map>
#include<queue>

using namespace std;
#define N 4
#define N2 16

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 space, MD, cost, f[N2];

  bool operator < (const Puzzle &p) const{
    for(int i = 0 ; i < N2 ; i++){
      if(f[i] == p.f[i]) continue;
      return f[i] < p.f[i];
    }
    return false;
  }
};

struct State{
  Puzzle puzzle;
  int estimated;
  bool operator < (const State &s) const{
    return estimated > s.estimated;
  }
};

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;
}

int astar(Puzzle s){
  priority_queue<State>PQ;
  s.MD = getAllMD(s);
  s.cost = 0;
  map<Puzzle, bool> V;
  Puzzle u, v;
  State initial;
  initial.puzzle = s;
  initial.estimated = getAllMD(s);
  PQ.push(initial);

  while(!PQ.empty()){
    State st = PQ.top();
    PQ.pop();
    u = st.puzzle;

    if(u.MD == 0) return u.cost;
    V[u] = true;

    int sx = u.space / N;
    int sy = u.space % N;

    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;
      v = u;

      v.MD -= MDT[tx * N + ty][v.f[tx * N + ty] - 1];
      v.MD += MDT[sx * N + sy][v.f[tx * N + ty] - 1];

      swap(v.f[sx * N + sy], v.f[tx * N + ty]);
      v.space = tx * N + ty;
      if(!V[v]){
        v.cost++;
        State news;
        news.puzzle = v;
        news.estimated = v.cost + v.MD;
        PQ.push(news);
      }
    }
  }

  return -1;
}


int main(){
  int i, j;

  for(i = 0 ; i < N2 ; i++)
    for(j = 0 ; j < N2 ; j++)
      MDT[i][j] = abs(i / N - j / N) + abs(i % N - j % N);

  Puzzle in;

  for(i = 0 ; i < N2 ; i++){
    cin >> in.f[i];
    if(in.f[i] == 0){
      in.f[i] = N2;
      in.space = i;
    }
  }

  cout<< astar(in)<<endl;

  return 0;
}