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_deepening(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 "unsolved";
}

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_deepening(in);
	cout << ans.size() << endl;

	return 0;
}