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<bits/stdc++.h>

static const int N = 4;
static const int N2 = 16;
static const int 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 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(std::max(prev, r) - std::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];
        std::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;

}

std::string iterative_deepning(Puzzle in){
    in.MD = getAllMD(in);

    for(limit = in.MD; limit <= LIMIT; limit++){
        state = in;
        if(dfs(0, -100)){
            std::string ans = "";
            for(int i = 0; i < limit; i++) ans += dir[path[i]];
            return ans;
        }
    }

    return "unsolvable";
}

int main(void){
    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++){
        std::cin >> in.f[i];
        if(in.f[i] == 0){
            in.f[i] = N2;
            in.space = i;
        }
    }

    std::string ans = iterative_deepning(in);
    std::cout << ans.size() << std::endl;

    return 0;
}