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

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 &pz) const
	{
		for (int i = 0; i < N2; ++i)
		{
			if (f[i] == pz.f[i]) continue;
			return f[i] < pz.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 = s.MD;
	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[tx * N + ty], v.f[sx * N + sy]);
			v.space  = tx * N + ty;
			if (!V[v])
			{
 				v.cost++;
				State news;
				news.puzzle = v;
				news.estimated = v.MD + v.cost;
				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;
		}

	}

	int ans = astar(in);
	cout<<ans<<endl;
	return 0;
}