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<stdio.h>
#include<iostream>
#include<cmath>
#include<string>
#include<cassert>
using namespace std;
#define N 4
#define N2 16
#define LIMIT 100

static const int dx[4] = { 0,-1,0,1 };
static const int dy[4] = { 1,0,-1,0 };
static const char dir[4] = { 'r','u','l','d' };
int MDT[N2][N2];

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

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

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

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

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[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_deeping(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]];
			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.f[i];
		if (in.f[i] == 0) {
			in.f[i] = N2;
			in.space = i;
		}
	}
	string ans = iterative_deeping(in);
	cout << ans.size() << endl;

	return 0;
}
