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 <iostream>
#include <cmath>
#include <map>
#include <queue>

using namespace std;
const int N = 4;
const int N2 = 16;

static const int dx[] = {0, -1, 0, 1};
static const int dy[] = {1, 0, -1, 0};
static const char dir[] = {'r', 'u', 'l', 'd'};
int MDT[N2][N2];

#define rep(i, n) for(int i=0; i<(int)n; ++i)

struct Puzzle{
    int f[N2], space, MD;
    int cost;

    bool operator<(const Puzzle &p) const{
        rep(i, N2){
            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;
    rep(i, N2){
        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()
{
    rep(i, N2)
        rep(j, N2)
            MDT[i][j] = abs(i % N - j % N) + abs(i / N - j / N);

    Puzzle in;

    rep(i, N2){
        cin >> in.f[i];
        if(in.f[i] == 0){
            in.f[i] = N2;
            in.space = i;
        }
    }
    cout << astar(in) << endl;
}