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<algorithm>
#include<queue>
#include<map>
#include<string>
using namespace std;
const int row = 4;
const int maxx = 16;
const int LIMIT = 100;
struct Puzzle {
	int f[maxx], space, MD;
};

const int dx[4] = { 0, -1, 0, 1 };
const int dy[4] = { 1, 0, -1 ,0 };
const char dir[4] = { 'r','u','l','d' };
int MDT[maxx][maxx];

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

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

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

bool dfs(int depth, int prev) {

	if (state.MD == 0)
		return true;

	// 如果当前的深度加上启发值超过了限制，则进行剪枝
	if (depth + state.MD > limit)
		return false;

	int sx = state.space / row;
	int sy = state.space % row;

	Puzzle tmp;
	for (int r = 0; r < 4; r++) {
		int tx = sx + dx[r];
		int ty = sy + dy[r];
		if (tx < 0 || ty < 0 || tx >= row || ty >= row)
			continue;
		if (max(prev, r) - min(prev, r) == 2)
			continue;
		tmp = state;
		// 计算曼哈顿距离的差值，同时交换拼图
		state.MD -= MDT[tx * row + ty][state.f[tx * row + ty] - 1];
		state.MD += MDT[sx * row + sy][state.f[tx * row + ty] - 1];
		swap(state.f[tx * row + ty], state.f[sx * row + sy]);
		state.space = tx * row + ty;
		if (dfs(depth + 1, r)) {
			path[depth] = r;
			return true;
		}
		state = tmp;
	}
	return false;
}

string iterative_deepening(Puzzle in) {
	// 初始状态的曼哈顿距离
	in.MD = getAllMD(in);
	for (limit = in.MD; limit <= LIMIT; limit++) {
		state = in;
		if (dfs(0, -100)) {
			string ans = "";
			for (int i = 0; i < limit; i++)
				ans += dir[path[i]];
			return ans;
		}
	}
	return "unsolvable";
}

int main()
{
	for (int i = 0; i < maxx; i++) {
		for (int j = 0; j < maxx; j++) {
			MDT[i][j] = abs(i / row - j / row) + abs(i % row - j % row);
		}
	}
	Puzzle in;
	for (int i = 0; i < maxx; i++) {
		cin >> in.f[i];
		if (in.f[i] == 0) {
			in.f[i] = maxx;
			in.space = i;
		}
	}
	string ans = iterative_deepening(in);
	printf("%d\n", ans.size());
	return 0;
}