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<cstdio>
#include<iostream>
#include<stack>
#include<vector>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#define _USE_MATH_DEFINES // M_PI????????????????????????
#include<math.h>
#include<queue>
#include<map>
#include<cassert>
#pragma warning(disable:4996)
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 iteratuve_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 "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 = iteratuve_deepening(in);
	cout << ans.size() << "\n";
	return 0;
}