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 int 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(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(in);
    cout << ans.size() << endl;
    
    return 0;
}
