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);
	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;	//f(n) = g(n) + h(n)
				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;
}