import sys sys.setrecursionlimit(10 ** 6) int1 = lambda x: int(x) - 1 p2D = lambda x: print(*x, sep="\n") def MI(): return map(int, sys.stdin.readline().split()) def LI(): return list(map(int, sys.stdin.readline().split())) def LLI(rows_number): return [LI() for _ in range(rows_number)] n, k = MI() l1, r1, l2, r2 = MI() md = 10 ** 9 + 7 def main(): # 1がIotが負けそう、2どちらも負けそう、3が髙橋が負けそう lose = [[[[0] * 5 for _ in range(5)] for _ in range(5)] for _ in range(5)] for il in range(5): for ir in range(5): if (il, ir) == (0, 0): continue for tl in range(5): for tr in range(5): if (tl, tr) == (0, 0): continue i_lose, t_lose = False, False if tl * tr == 0 and (5 - tl - tr) % 5 in [il, ir]: t_lose = True if il * ir == 0 and (5 - il - ir) % 5 in [tl, tr]: i_lose = True if i_lose and t_lose: lose[il][ir][tl][tr] = 2 elif i_lose: lose[il][ir][tl][tr] = 1 elif t_lose: lose[il][ir][tl][tr] = 3 # i=0がItoの番、1が髙橋の番で遷移先のリスト to = [[[[[[], [], [], [], []] for _ in range(5)] for _ in range(5)] for _ in range(5)] for _ in range(2)] # 遷移先リストを作る for i in range(1, k + 1): for il in range(5): for ir in range(5): if (il, ir) == (0, 0): continue for tl in range(5): for tr in range(5): if (tl, tr) == (0, 0): continue if i % 2: # Iotのターン to_value = [] # 攻撃 if il > 0 and tl > 0: to_value.append((il, ir, (tl + il) % 5, tr)) if il > 0 and tr > 0: to_value.append((il, ir, tl, (tr + il) % 5)) if ir > 0 and tl > 0: to_value.append((il, ir, (tl + ir) % 5, tr)) if ir > 0 and tr > 0: to_value.append((il, ir, tl, (tr + ir) % 5)) # 分割(自滅、交換、重複がないように) ng = set([(0, 0), (il, ir), (ir, il)]) for nl in range(il + ir + 1): nr = il + ir - nl nl, nr = nl % 5, nr % 5 if (nl, nr) in ng: continue ng.add((nl, nr)) to_value.append((nl, nr, tl, tr)) to[0][il][ir][tl][tr] = to_value else: # 髙橋のターン # 勝てるときは必ず勝つ if 0 < lose[il][ir][tl][tr] < 3: continue to_value = [] # 詰んでいるとき負けパターン数とそのフラグ to_lose = [] flag_lose = True # 攻撃 if il > 0 and tl > 0: nl, nr = (il + tl) % 5, ir pre = (nl, nr, tl, tr) to_lose.append(pre) if lose[nl][nr][tl][tr] < 2: flag_lose = False to_value.append(pre) if il > 0 and tr > 0: nl, nr = (il + tr) % 5, ir pre = (nl, nr, tl, tr) to_lose.append(pre) if lose[nl][nr][tl][tr] < 2: flag_lose = False to_value.append(pre) if ir > 0 and tl > 0: nl, nr = il, (ir + tl) % 5 pre = (nl, nr, tl, tr) to_lose.append(pre) if lose[nl][nr][tl][tr] < 2: flag_lose = False to_value.append(pre) if ir > 0 and tr > 0: nl, nr = il, (ir + tr) % 5 pre = (nl, nr, tl, tr) to_lose.append(pre) if lose[nl][nr][tl][tr] < 2: flag_lose = False to_value.append(pre) # 分割(自滅、交換、重複がないように) ng = set([(0, 0), (tl, tr), (tr, tl)]) for nl in range(tl + tr + 1): nr = tl + tr - nl nl, nr = nl % 5, nr % 5 if (nl, nr) in ng: continue ng.add((nl, nr)) pre = (il, ir, nl, nr) to_lose.append(pre) if lose[il][ir][nl][nr] < 2: flag_lose = False to_value.append(pre) to[1][il][ir][tl][tr] = to_lose if flag_lose else to_value # 手がil,ir,tl,trの状態からiターンでIotが勝つ方法を # dp[i][il][ir][tl][tr]とする dp = [[[[0] * 5 for _ in range(5)] for _ in range(5)] for _ in range(5)] # 初期化 for il in range(5): for ir in range(5): if (il, ir) == (0, 0): continue dp[il][ir][0][0] = 1 ans = 0 for i in range(k): pdp = dp dp = [[[[0] * 5 for _ in range(5)] for _ in range(5)] for _ in range(5)] for il in range(5): for ir in range(5): if (il, ir) == (0, 0): continue for tl in range(5): for tr in range(5): if (tl, tr) == (0, 0): continue dp_value = 0 for pil, pir, ptl, ptr in to[i % 2][il][ir][tl][tr]: dp_value += pdp[pil][pir][ptl][ptr] dp[il][ir][tl][tr] = dp_value % md if i % 2 == n: ans += dp[l1][r1][l2][r2] ans %= md print(ans) main()