# コードを綺麗にした from math import gcd from collections import defaultdict mod = 998244353 # m ... 決まっている鯨の数 # rl ... 方向の個数 (長さ 4 の配列) # x ... 鯨の x 座標 (長さ m の配列) # y ... 鯨の y 座標 (長さ m の配列) # r ... 鯨の方向 (長さ m の配列) # w ... 盤面の x 軸の広さ # h ... 盤面の y 軸の広さ # g ... gcd(w, h) # TYPE 1 縦・横のみ def count_type1(m, rl, x, y, r, w, h, g): ret = 0 # 横 if rl[0] + rl[1] == 0: d = defaultdict(lambda: -1) mode = 1 for i in range(m): if d[y[i]] == -1: d[y[i]] = r[i] elif d[y[i]] != r[i]: mode = 0 break if mode: f = len(d) ret += pow(2, h-f, mod) ret %= mod # 縦 if rl[2] + rl[3] == 0: d = defaultdict(lambda: -1) mode = 1 for i in range(m): if d[x[i]] == -1: d[x[i]] = r[i] elif d[x[i]] != r[i]: mode = 0 break if mode: f = len(d) ret += pow(2, w-f, mod) ret %= mod return ret # TYPE 2 偶数自明 (g = 0 mod 2) def count_type2(m, rl, x, y, r, w, h, g): ret = 0 if g % 2 == 0: # パリティで場合分けする for parity in range(2): d1 = defaultdict(lambda: -1) d2 = defaultdict(lambda: -1) mode = 1 for i in range(m): if r[i] == 0 or r[i] == 1: # 縦 if (x[i] + y[i]) % 2 != parity: mode = 0 break if d1[x[i]] == -1: d1[x[i]] = r[i] elif d1[x[i]] != r[i]: mode = 0 break else: # 横 if (x[i] + y[i]) % 2 != 1 ^ parity: mode = 0 break if d2[y[i]] == -1: d2[y[i]] = r[i] elif d2[y[i]] != r[i]: mode = 0 break if mode: ret += pow(2, w-len(d1) + h-len(d2), mod) ret %= mod return ret # TYPE 3 (x-y)%g で場合分け def count_type3(m, rl, x, y, r, w, h, g): ret = 0 for v in [(1, 3), (0, 2)]: if rl[v[0]] == rl[v[1]] == 0: mode = 1 d = defaultdict(lambda:-1) for i in range(m): t = (x[i] - y[i]) % g if d[t] == -1: d[t] = r[i] elif d[t] != r[i]: mode = 0 break if mode: ret += pow(2, g-len(d), mod) ret %= mod return ret # TYPE 4 (x+y)%g で場合分け def count_type4(m, rl, x, y, r, w, h, g): ret = 0 for v in [(1, 2), (0, 3)]: if rl[v[0]] == rl[v[1]] == 0: mode = 1 d = defaultdict(lambda:-1) for i in range(m): t = (x[i] + y[i]) % g if d[t] == -1: d[t] = r[i] elif d[t] != r[i]: mode = 0 break if mode: ret += pow(2, g-len(d), mod) ret %= mod return ret # TYPE 4.5.1 # TYPE 3, 4 のうち, TYPE 1 にあてはまるものを除く def count_type451(m, rl, x, y, r, w, h, g): ret = 0 for v in range(4): if rl[0] + rl[1] + rl[2] + rl[3] - rl[v] == 0: ret -= 2 ret %= mod return ret # TYPE 4.5.2 # TYPE 3, 4 のうち, TYPE 2 にあてはまるものを除く def count_type452(m, rl, x, y, r, w, h, g): ret = 0 if g % 2 == 0: for p in range(4): for q in range(4): if p//2 == q//2: continue mode = 1 for i in range(m): t = (x[i] + y[i]) % 2 if t == 0 and r[i] != p: mode = 0 break if t == 1 and r[i] != q: mode = 0 break if mode: ret -= 1 ret %= mod return ret # 非自明の 48 通りを作る def make_hijimei(): ret = [] for num in range(4 ** 8): v = num a = [[0] * 4 for i in range(4)] rl = [0] * 4 for i in range(8): t = v % 4 v //= 4 a[i//4][i%4] = t a[i//4 + 2][(i+2)%4] = t rl[t] += 2 # 衝突するか否か mode = 1 for i in range(4): md = -1 for j in range(4): if a[i][j] == 0 or a[i][j] == 1: if md == -1: md = a[i][j] elif md != a[i][j]: mode = 0 break for j in range(4): md = -1 for i in range(4): if a[i][j] == 2 or a[i][j] == 3: if md == -1: md = a[i][j] elif md != a[i][j]: mode = 0 break if mode == 0: continue b = [[0] * 4 for i in range(4)] for i in range(4): for j in range(4): b[i][j] = a[i][j] # 愚直にシミュレーションする for tyr in range(20): c = [[-1] * 4 for i in range(4)] for i in range(4): for j in range(4): if b[i][j] == 0: if c[i][(j+1)%4] != -1: mode = 0 break c[i][(j+1)%4] = b[i][j] if b[i][j] == 1: if c[i][(j-1)%4] != -1: mode = 0 break c[i][(j-1)%4] = b[i][j] if b[i][j] == 2: if c[(i-1)%4][j] != -1: mode = 0 break c[(i-1)%4][j] = b[i][j] if b[i][j] == 3: if c[(i+1)%4][j] != -1: mode = 0 break c[(i+1)%4][j] = b[i][j] if mode == 0: break b = c # 衝突しなかった場合, TYPE 1, 2, 3, 4 に当てはまるかを確認 if mode: ans = 0 m = 16 x = [0] * m y = [0] * m r = [0] * m rl = [0] * 4 for i in range(4): for j in range(4): x[i*4 + j] = i y[i*4 + j] = j r[i*4 + j] = a[i][j] rl[a[i][j]] += 1 w = 4 h = 4 g = 4 ans += count_type1(m, rl, x, y, r, w, h, g) ans += count_type2(m, rl, x, y, r, w, h, g) ans += count_type3(m, rl, x, y, r, w, h, g) ans += count_type4(m, rl, x, y, r, w, h, g) ans += count_type451(m, rl, x, y, r, w, h, g) ans += count_type452(m, rl, x, y, r, w, h, g) # TYPE 1, 2, 3, 4 に当てはまらない場合, 非自明 if ans == 0: ret.append(a) return ret # 実行! ar = make_hijimei() # TYPE 5 # 非自明 48通りを全探索する def count_type5(m, rl, x, y, r, w, h, g): ret = 0 if g % 4 == 0: a = [[-1] * 4 for i in range(4)] mode = 1 for i in range(m): if a[x[i]%4][y[i]%4] == -1: a[x[i]%4][y[i]%4] = r[i] elif a[x[i]%4][y[i]%4] != r[i]: mode = 0 break if mode == 0: return 0 for b in ar: mode = 1 for i in range(4): for j in range(4): if a[i][j] != -1 and a[i][j] != b[i][j]: mode = 0 break if mode: ret += 1 return ret # 解く def solve(): # 順番逆のほうが直感的なので逆にしてます(ごめんね) w, h, m = map(int,input().split()) v = {"U":0, "D":1, "L":2, "R":3} g = gcd(w, h) rl = [0] * 4 # 方向のカウント x = [0] * m # X y = [0] * m # Y r = [0] * m # 方向 # 入力 for i in range(m): xs, ys, d = input().split() x[i] = int(xs) y[i] = int(ys) r[i] = v[d] rl[r[i]] += 1 # 数え上げ ans = 0 ans += count_type1(m, rl, x, y, r, w, h, g) ans += count_type2(m, rl, x, y, r, w, h, g) ans += count_type3(m, rl, x, y, r, w, h, g) ans += count_type4(m, rl, x, y, r, w, h, g) ans += count_type451(m, rl, x, y, r, w, h, g) ans += count_type452(m, rl, x, y, r, w, h, g) ans += count_type5(m, rl, x, y, r, w, h, g) ans %= mod return ans T = int(input()) for _ in range(T): print(solve())