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

using namespace std;
#define N 4
#define N2 16

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;
    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 aster(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 << aster(in) << endl;
    
    return 0;
}
