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 <string>
#include <cassert>
using namespace std;

const long N = 4;
const long N2 = 16;
const long LIMIT = 100;
const long dx[4] = {0, -1, 0, 1};
const long dy[4] = {1, 0, -1, 0};
const char dir[4] = {'r', 'u', 'l', 'd'};
long MDT[N2][N2];

struct Puzzle {long f[N2], space, MD;};

Puzzle state;
long limit;
long path[LIMIT];

long getAllMD(Puzzle pz){
	long sum = 0;
	for (long i = 0; i < N2; i++){
		if (pz.f[i] == N2) continue;
		sum += MDT[i][pz.f[i] - 1];
	}
	return sum;
}

bool isSolved(){
	for (long i = 0; i < N2; i++){
		if (state.f[i] != i + 1) return false;
	}
	return true;
}

bool dfs(long depth, long prev){
	if (state.MD == 0) return true;
	if (depth + state.MD > limit) return false;

	long sx = state.space / N;
	long sy = state.space % N;
	Puzzle tmp;

	for (long r = 0; r < 4; r++){
		long tx = sx + dx[r];
		long ty = sy + dy[r];
		if (tx < 0 || ty < 0 || tx >= N || ty >= N) continue;
		if (max(prev, r) - min(prev, r) == 2) continue;
		tmp = state;

		state.MD -= MDT[tx * N + ty][state.f[tx * N + ty] - 1];
		state.MD += MDT[sx * N + sy][state.f[tx * N + ty] - 1];
		swap(state.f[tx * N + ty], state.f[sx * N + sy]);
		state.space = tx * N + ty;
		if (dfs(depth + 1, r)){
			path[depth] = r;
			return true;
		}
		state = tmp;
	}
	return false;
}

string iterative_deepening(Puzzle in){
	in.MD = getAllMD(in);

	for (limit = in.MD; limit <= LIMIT; limit++){
		state = in;
		if (dfs(0, -100)){
			string ans = "";
			for (long i = 0; i < limit; i++) ans += dir[path[i]];
			return ans;
		}
	}
	return "unsolvable";
}

int main(){
	for (long i = 0; i < N2; i++){
		for (long j = 0; j < N2; j++){
			MDT[i][j] = abs(i / N - j / N) + abs(i % N - j % N);
			Puzzle in;
		}
	}
	Puzzle in;
	for (long i = 0; i < N2; i++){
		cin >> in.f[i];
		if (in.f[i] == 0){
			in.f[i] = N2;
			in.space = i;
		}
	}
	string ans = iterative_deepening(in);
	cout << ans.size() << endl;

	return 0;
}