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 <string>
#include <algorithm>
using namespace std;
#define N 4
#define N2 16
#define 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 isSolve(Puzzle s) {
    for (int i = 0; i != N2; ++i) {
        if ( s.f[i] != i + 1 ) return false;
    }
    return true;
}

bool dfs(int depth, int prev) {
    if ( state.MD == 0 ) return true;
    if ( state.MD + depth > limit ) return false;

    Puzzle tmp;
    int sx = state.space / N;
    int sy = state.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;
        if ( max(r, prev) - min (r, prev) == 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];
        swap(state.f[sx * N + sy], state.f[tx * N + ty]);
        state.space = tx * N + ty;
        if ( dfs(depth + 1, r) ) {
            path[depth] = r;
            return true;
        }
        state = tmp;
    }
    return false;
}

string iterative_deepening(Puzzle in) {
    in.MD = getAllMD(in);

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

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;
        }
    }
    string ans = iterative_deepening(in);
    cout << ans.size() << endl;
    return 0;
}