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<cstdio>
#include<map>
#include<queue>
#include<cmath>
using namespace std;

#define N 4
#define N2 16

const int dx[4] = {0, -1, 0, 1};
const int dy[4] = {1, 0, -1, 0};

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;
	}
};

int getallMD(const Puzzle &p)
{
	int h = 0;
	for(int i = 0; i < N2; i++){
	if(i == p.space) continue;
		h += MDT[i][p.f[i] - 1];
	}
	return h;
}

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

int astar(Puzzle p)
{
	priority_queue<State> A;
	map<Puzzle, bool> USED;

	State initial;
	initial.pz = p;
	initial.estimated = p.MD;
	A.push(initial);
	USED[p] = true;

	Puzzle u, v;
	int sx, sy, tx, ty, r;

	while(!A.empty())
	{
		u = A.top().pz; A.pop();
		if(u.MD == 0) return u.cost;

		sx = u.space / N;
		sy = u.space % N;
		for(r = 0; r < 4; r++){
			tx = sx + dx[r];
			ty = sy + dy[r];
			if(tx < 0 || tx >= N || ty < 0 || ty >= N) continue;
			v = u;
			v.MD -= MDT[tx * N + ty][u.f[tx * N + ty] - 1];
			v.MD += MDT[sx * N + sy][u.f[tx * N + ty] - 1];
			swap(v.f[sx * N + sy], v.f[tx * N + ty]);
			v.space = tx * N + ty;
			
			if(!USED[v]){
				v.cost++;
				USED[v] = true;
				State new_s;
				new_s.pz = v;
				new_s.estimated = v.cost + v.MD;
				A.push(new_s);
			}
		}
	};
	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++){
		scanf("%d", &in.f[i]);
		if(in.f[i] == 0) in.space = i;
	}
	in.cost = 0;
	in.MD = getallMD(in);

	cout << astar(in) << endl;

	return 0;
}