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
#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;
	int cost;
	bool operator < (const Puzzle &p) const {
		for (int i = 0; i < maxx; i++) {
			if (f[i] != p.f[i])
				return f[i] < p.f[i];
		}
		return false;
	}
};

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


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;
}

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 / row;
		int sy = u.space % row;
		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;
			v = u;
			// 计算曼哈顿距离的差值，同时交换拼图
			v.MD -= MDT[tx * row + ty][v.f[tx * row + ty] - 1];
			v.MD += MDT[sx * row + sy][v.f[tx * row + ty] - 1];

			swap(v.f[tx * row + ty], v.f[sx * row + sy]);
			v.space = tx * row + ty;
			if (!V[v]) {
				v.cost++;
				State news;
				news.puzzle = v;
				news.estimated = v.cost + v.MD;
				PQ.push(news);
			}
		}
	}

	return -1;
}


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;
		}
	}

	printf("%d\n", astar(in));
	return 0;
}
