#!/bin/python3

import sys

def printShortestPath(n, i_start, j_start, i_end, j_end):
    #  Print the distance along with the sequence of moves.
    # Run a bfs
    level = [(i_start, j_start)]
    move = {}
    parent = {(i_start, j_start): None}
    #options = {'UL': (-2, -1), 'UR': (-2, 1), 'L': (0, -2), 'R': (0, 2), 'LL': (2, -1), 'LR': (2, 1)}
    options = [('UL', (-2, -1)), ('UR', (-2, 1)), ('R', (0, 2)), ('LR', (2, 1)), ('LL', (2, -1)), ('L', (0, -2))]
    #UL, UR, R, LR, LL, L
    n_level = 0
    while level:
        new_level = []
        for i, j in level:
            if i == i_end and j == j_end:
                # print the output
                print(n_level)
                moves = []
                curr = (i_end, j_end)
                while curr in move:
                    moves.append(move[curr])
                    curr = parent[curr]
                print(' '.join(moves[::-1]))
                return None
            else:
                # Push all unvisited neighbors
                for key, value in options:
                    di, dj = value
                    new_i, new_j = i + di, j + dj
                    if 0 <= new_i < n and 0 <= new_j < n and (new_i, new_j) not in parent:
                        new_level.append((new_i, new_j))
                        parent[(new_i, new_j)] = (i, j)
                        move[(new_i, new_j)] = key
        level = new_level
        n_level += 1
    print('Impossible')

if __name__ == "__main__":
    n = int(input().strip())
    i_start, j_start, i_end, j_end = input().strip().split(' ')
    i_start, j_start, i_end, j_end = [int(i_start), int(j_start), int(i_end), int(j_end)]
    printShortestPath(n, i_start, j_start, i_end, j_end)