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:
// Iterative Deepening
#include <cstdio>
#include <cmath>
#include <string>
#include <cassert>
using namespace std;

#define N (4)
#define N2 (N * N)
#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];
    int space;
    int 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 "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++ ) {
	scanf("%d", &in.f[i]);
	if ( in.f[i] == 0 ) {
	    in.f[i] = N2;
	    in.space = i;
	}
    }

    string ans = iterative_deepening( in );

    printf("%d\n", (int)ans.size());

    return 0;
}