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:
//
// Created by ?????°?????? on 2016/07/02.
//

#include<iostream>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
using namespace std;

static const int N = 4;
static const int N2 =16;

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;
    int cost;

    bool operator < (const Puzzle &p) const{
        for (int i = 0; i < N2 ; ++i) {
            if(f[i] == p.f[i]) continue;
            return f[i] < p.f[i];
        }
        return false;
    }
};

struct State{
    Puzzle puzzle;
    int estimated;
    bool operator < (const State &s) const{
        return estimated > s.estimated;
    }
};

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;
}

int astar(Puzzle s){
    priority_queue<State> PQ;
    s.MD = getALLMD(s);
    s.cost = 0;
    map<Puzzle,bool> V;
    Puzzle u,v;
    State initial;
    initial.puzzle = s;
    initial.estimated = getALLMD(s);
    PQ.push(initial);

    while (!PQ.empty()){
        State st = PQ.top(); PQ.pop();
        u = st.puzzle;

        if (u.MD == 0) return u.cost;
        V[u] = true;

        int sx = u.space / N;
        int sy = u.space % N;

        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;
            v = u;

            v.MD -= MDT[tx * N + ty][v.f[tx * N +ty] - 1];
            v.MD += MDT[sx * N + sy][v.f[tx * N +ty] - 1];

            swap(v.f[sx * N + sy],v.f[tx * N + ty]);
            v.space = tx * N + ty;
            if(!V[v]){
                v.cost ++;
                State news;
                news.puzzle = v;
                news.estimated = v.cost + v.MD;
                PQ.push(news);
            }
        }
    }
    return -1;
}

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;
        }
    }
    cout << astar(in) << endl;
    return 0;
}