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 <bits/stdc++.h>
using namespace std;

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[16][16];

struct Puzzle{
    int f[16];
    int space;
    int MD;
};

Puzzle state;
int limit;
int path[100];

int getAllMD(Puzzle &pz){
    int sum=0;
    for(int i=0;i<16;i++){
        if(pz.f[i]==16) continue;
        sum+=MDT[i][pz.f[i]-1];
    }
    return sum;
}

bool dfs(int depth,int prev){
    if(state.MD==0) return true;
    if(depth+state.MD>limit) return false;

    int sx=state.space/4;
    int sy=state.space%4;
    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>3||ty>3){
            continue;
        }
        if(abs(prev-r)==2){
            continue;
        }
        tmp=state;
        state.MD-=MDT[tx*4+ty][state.f[tx*4+ty]-1];
        state.MD+=MDT[sx*4+sy][state.f[tx*4+ty]-1];
        swap(state.f[state.space],state.f[tx*4+ty]);
        state.space=tx*4+ty;
        if(dfs(depth+1,r)){
            path[depth]=r;
            return true;
        }
        state=tmp;
    }
    return false;
}

string iterative_deepening(Puzzle x){
    x.MD=getAllMD(x);
    for(limit=x.MD;limit<100;limit++){
        state=x;
        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<16;i++){
        for(int j=0;j<16;j++){
            MDT[i][j]=abs(i/4-j/4)+abs(i%4-j%4);
        }
    }
    Puzzle x;
    for(int i=0;i<16;i++){
        cin>>x.f[i];
        if(x.f[i]==0){
            x.f[i]=16;
            x.space=i;
        }
    }

    cout<<iterative_deepening(x).size()<<endl;
}
