Problem:
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15 Puzzle

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.

Input

The $4 \times 4$ integers denoting the pieces or space are given.

Output

Print the fewest steps in a line.

Constraints

The given puzzle is solvable in at most 45 steps.

Sample Input

1 2 3 4
6 7 8 0
5 10 11 12
9 13 14 15

Sample Output

8

Solution in Python3 (judge verdict: Accepted)

import sys
from heapq import heappop, heappush


adjacent = (
    (1, 4),
    (0, 2, 5),
    (1, 3, 6),
    (2, 7),
    (0, 5, 8),
    (1, 4, 6, 9),
    (2, 5, 7, 10),
    (3, 6, 11),
    (4, 9, 12),
    (5, 8, 10, 13),
    (6, 9, 11, 14),
    (7, 10, 15),
    (8, 13),
    (9, 12, 14),
    (10, 13, 15),
    (11, 14)
)

distance = (
    (),
    (0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5, 6),
    (1, 0, 1, 2, 2, 1, 2, 3, 3, 2, 3, 4, 4, 3, 4, 5),
    (2, 1, 0, 1, 3, 2, 1, 2, 4, 3, 2, 3, 5, 4, 3, 4),
    (3, 2, 1, 0, 4, 3, 2, 1, 5, 4, 3, 2, 6, 5, 4, 3),
    (1, 2, 3, 4, 0, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 5),
    (2, 1, 2, 3, 1, 0, 1, 2, 2, 1, 2, 3, 3, 2, 3, 4),
    (3, 2, 1, 2, 2, 1, 0, 1, 3, 2, 1, 2, 4, 3, 2, 3),
    (4, 3, 2, 1, 3, 2, 1, 0, 4, 3, 2, 1, 5, 4, 3, 2),
    (2, 3, 4, 5, 1, 2, 3, 4, 0, 1, 2, 3, 1, 2, 3, 4),
    (3, 2, 3, 4, 2, 1, 2, 3, 1, 0, 1, 2, 2, 1, 2, 3),
    (4, 3, 2, 3, 3, 2, 1, 2, 2, 1, 0, 1, 3, 2, 1, 2),
    (5, 4, 3, 2, 4, 3, 2, 1, 3, 2, 1, 0, 4, 3, 2, 1),
    (3, 4, 5, 6, 2, 3, 4, 5, 1, 2, 3, 4, 0, 1, 2, 3),
    (4, 3, 4, 5, 3, 2, 3, 4, 2, 1, 2, 3, 1, 0, 1, 2),
    (5, 4, 3, 4, 4, 3, 2, 3, 3, 2, 1, 2, 2, 1, 0, 1)
)

end = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 0]


def get_distance():
    global board
    sum_distance = 0
    for i in range(16):
        if board[i] == 0:
            continue
        sum_distance += distance[board[i]][i]
    return sum_distance


def search(limit, move, space, lower, prev_space):
    global board

    if move == limit:
        if board == end:
            return True

    else:
        for next_space in adjacent[space]:
            if next_space == prev_space:
                continue
            board[space] = board[next_space]
            board[next_space] = 0

            new_lower = lower - distance[board[space]][next_space] + distance[board[space]][space]

            if new_lower + move <= limit:
                if search(limit, move + 1, next_space, new_lower, space):
                    return True
            board[next_space] = board[space]
            board[space] = 0


board = []
for _ in range(4):
    board.extend(list(map(int, sys.stdin.readline().strip().split())))

# 最初のスペースの位置で、解の手順が奇数手か偶数手かが決定される
n = get_distance()  # 暫定的な最小手
first_space = board.index(0)
e_o = (0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0)
if n % 2 != e_o[first_space]:
    n += 1

for l in range(n, 46, 2):
    if search(l, 0, first_space, n, None):
        print(l)
        break