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 f[N2], space, MD;
  int cost;
 
  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) + s.cost  ;
  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() {
  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;
    }
  }
  cout << astar(in) << endl;
 
  return 0;
}