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 <algorithm>
#include <cstdio>
#include <cstdlib>
#include <iostream>
#include <string>

using namespace std;
#define N 4
#define N2 16
#define LIMIT 45
const int dx[4] { -1, 0, 1, 0 };
const int dy[4] { 0, -1, 0, 1 };
const char dir[4] { 'u', 'l', 'd', 'r' };
struct Puzzle {
	int f[N2], space, MD;
};
int MDT[N2][N2];
int getAllMd(Puzzle p) {
	int sum = 0;
	for (int i = 0; i < N2; ++i) {
		if (p.f[i] == N2)
			continue;
		sum += MDT[i][p.f[i] - 1];
	}
	return sum;
}

int limit;
int path[LIMIT];
Puzzle state;
bool dfs(int depth, int prev) {

	if (state.MD == 0)
		return true;
	if (depth + state.MD > limit)
		return false;

	int sx = state.space / N;
	int sy = state.space % N;
	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 >= 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[sx * N + sy], state.f[tx * N + ty]);
		state.space = tx * N + ty;

		if (dfs(depth + 1, r)) {
			path[depth] = r;
			return true;
		}
		state = tmp;
	}
	return false;
}
string deepening(Puzzle p) {
	int md = getAllMd(p);
	p.MD = md;
	state = p;
	for (limit = md; limit <= LIMIT; ++limit) {
		string ans = "";
		if (dfs(0, -100)) {
			for (int i = 0; i < limit; ++i) {
				ans += dir[path[i]];
			}
			return ans;
		}
	}
	return "";
}

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 p;
	for (int i = 0; i < N2; ++i) {
		scanf("%d", &p.f[i]);
		if (p.f[i] == 0) {
			p.f[i] = N2;
			p.space = i;
		}
	}

	string ans = deepening(p);
	cout << ans.size() << endl;

	return 0;
}