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>
using namespace std;

int MDT[16][16];
const int nxt[][2] = {{0,1},{-1,0},{0,-1},{1,0} };
const char dir[] = {'r','u','l','d'};

struct Puzzle {
    int f[16],space,MD;
    int cost;

    bool operator < (const Puzzle& p) const {
        for(int i = 0; i < 16; 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 < 16; i++) {
        if(pz.f[i] == 16) 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;
    State init;
    init.puzzle = s;
    init.estimated = getAllMD(s);
    PQ.push(init);
    while(!PQ.empty()) {
        State st = PQ.top(); PQ.pop();
        Puzzle u = st.puzzle;
        if(u.MD == 0) return u.cost;
        V[u] = true;
        int sx = u.space / 4;
        int sy = u.space % 4;
        for(int i = 0; i < 4; i++) {
            int tx = sx + nxt[i][0];
            int ty = sy + nxt[i][1];
            if(tx < 0 || ty < 0 || tx >= 4 || ty >= 4) continue;
            Puzzle v = u;
            swap(v.f[sx*4+sy],v.f[tx*4+ty]);
            v.space = tx * 4 + ty;
            v.MD = getAllMD(v);
            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 < 16; i++) {
        for(int j = 0; j < 16; j++) {
            MDT[i][j] = abs(i/4-j/4) + abs(i%4-j%4);
        } // 计算i位置到j位置的曼哈顿距离
    } // i位置就是[i/4][i%4],j位置以此类推
    Puzzle in; //初始拼图
    for(int i = 0; i < 16; i++) {
        cin >> in.f[i];
        if(in.f[i] == 0) {
            in.f[i] = 16;
            in.space = i;
        }
    }
    cout << Astar(in) << endl; // A*搜索
    return 0;
}