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<algorithm>
#include<queue>
#include<string>
#include<map>
using namespace std;

static const int N = 4;
static const int N2 = 16;
static const int LIMIT = 100;

static const int dx[4] = {-1, 0, 1, 0}; //??£??\???{}
static const int dy[4] = {0, -1, 0, 1};
static const int dir[4] = {'u', 'l', 'd', 'r'};

int MDT[N2][N2]; //???????????????????????¢

struct Puzzle {
	int p[N2], space, MD;
	string path;

	bool operator < (const Puzzle &p1) const { //????????§?????????
		for (int i = 0; i < N2; i++) {
			if (p[i] == p1.p[i]) continue;
			return p[i] > p1.p[i];
		}
		return false;
	}
};

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

int getAllMD(Puzzle pz) {
	int sum = 0;
	for (int i = 0; i < N2; i++) {
		if (pz.p[i] == N2) continue;
		sum += MDT[i][pz.p[i] - 1]; //i????????¨????????????j???????????????
	}
	return sum;
}

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; //??¢?´¢???????????£??????????????????????????????prev
		tmp = state;
		state.MD -= MDT[tx * N + ty][state.p[tx * N + ty] - 1];
		state.MD += MDT[sx * N + sy][state.p[tx * N + ty] - 1];
		swap(state.p[tx * N + ty], state.p[sx * N + sy]);
		state.space = tx * N + ty;
		if (dfs(depth + 1, r)) {
			path[depth] = r;
			return true;
		}
		state = tmp; //??±??????????????????
	}
	return false;
}

string ida(Puzzle in) {
	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]]; //????°???????limit??¨??????????????°?§??????????
			return ans;
		}
	}
	return "unsolvable";
}

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.p[i];
		if (!in.p[i]) {
			in.p[i] = N2; //isTarget?????????
			in.space = i;
		}
	}

	string ans = ida(in);
	cout << ans.size() << endl;

	return 0;
}