# this code was generated by grok 3 beta (x.com)

def inverse_permutation(perm):
    # 順列の逆順列を計算（1-based）
    n = len(perm)
    inv = [0] * n
    for i in range(n):
        inv[perm[i] - 1] = i + 1
    return inv

def apply_row_perm(matrix, row_idx, perm):
    # 行を順列に基づいて並び替える（行のインデックスを変更）
    n = len(matrix)
    new_matrix = [None] * n
    for i in range(n):
        new_matrix[i] = matrix[perm[i] - 1][:]  # 行 perm[i] を i 番目に
    return new_matrix

def apply_col_perm(matrix, col_idx, perm):
    # 列を順列に基づいて並び替える（列のインデックスを変更）
    n = len(matrix)
    new_matrix = [[0] * n for _ in range(n)]
    for j in range(n):
        for i in range(n):
            new_matrix[i][j] = matrix[i][perm[j] - 1]  # 列 perm[j] を j 番目に
    return new_matrix

def check_equal(matrix1, matrix2):
    # 2つの行列が等しいか確認
    return all(row1 == row2 for row1, row2 in zip(matrix1, matrix2))

def dfs(current, A, operations, limit, N):
    # バックトラックで B から A への操作列を探索
    if check_equal(current, A):
        return operations
    if len(operations) >= limit:
        return None

    # 各行を試す
    for i in range(N):
        perm = current[i]
        inv_perm = inverse_permutation(perm)
        new_matrix = apply_row_perm(current, i, inv_perm)
        new_ops = operations + [("R", i + 1)]
        result = dfs(new_matrix, A, new_ops, limit, N)
        if result is not None:
            return result

    # 各列を試す
    for j in range(N):
        perm = [current[i][j] for i in range(N)]
        inv_perm = inverse_permutation(perm)
        new_matrix = apply_col_perm(current, j, inv_perm)
        new_ops = operations + [("C", j + 1)]
        result = dfs(new_matrix, A, new_ops, limit, N)
        if result is not None:
            return result

    return None

def solve(N, K, A, B):
    # B から A への操作列をバックトラックで探索
    operations = []
    limit = 1000  # 操作回数の上限
    result = dfs(B, A, operations, limit, N)
    if result is None:
        return []  # 解が見つからない場合（問題保証により発生しない）
    return result

# 入力処理
N, K = map(int, input().split())
A = [list(map(int, input().split())) for _ in range(N)]
B = [list(map(int, input().split())) for _ in range(N)]

# 解を求める
ops = solve(N, K, A, B)

# 出力
print(len(ops))
for op_type, pos in ops:
    print(f"{op_type} {pos}")