# 問題の首座として、通れない、としたときにどうなるか、を高速で求めたい。
# 通れないは難しいから、通るものを数えて、全体から引く
# 全体から引くときに、各々引いて、,,,とやるとかぶるから包除原理
import sys

input = sys.stdin.buffer.readline
mod = 10 ** 8 + 7
h, w, k = map(int, input().split())

AB = [list(map(int, input().split())) for i in range(k)]

# AB.sort(key=lambda x: [x[1], x[2]])
# idx = [i for i, _, __ in AB]

N = h + w + 100 + 5
bikkuri = [0] * N
bikkuri[0] = 1
gyaku = [0] * N
for i in range(1, N):
    bikkuri[i] = (i * bikkuri[i - 1]) % mod
gyaku[N - 1] = pow(bikkuri[N - 1], mod - 2, mod)
for i in range(N)[::-1]:
    gyaku[i - 1] = (gyaku[i] * i) % mod


def comb(n, r):
    if r < 0 or n - r < 0:
        return 0
    return bikkuri[n] * gyaku[r] % mod * gyaku[n - r] % mod


popcount = [0] * (1 << k)

for i in range(1 << k):
    popcount[i] = popcount[i // 2] + i % 2


memo = [0] * (1 << k)

for i in range(1 << k):
    temp = 1
    kouho = []
    for j in range(k):
        if (i >> j) & 1:
            kouho.append(AB[j])
    kouho.sort()
    px, py = 0, 0

    for nx, ny in kouho:
        dx = nx - px
        dy = ny - py
        temp *= comb(dx + dy, dy)
        temp %= mod
        px, py = nx, ny
    nx, ny = h, w

    dx = nx - px
    dy = ny - py

    temp *= comb(dx + dy, dy)
    temp %= mod
    memo[i] = temp

total = comb(h + w, w)


ans = [0] * (1 << k)

for i in range(1 << k):
    j = i
    while j:
        # print("i,j", i, j)
        keisu = 1 if popcount[j] % 2 else -1
        ans[i] += keisu * memo[j]
        ans[i] %= mod
        # print(keisu * memo[j])
        j -= 1
        j &= i

ans = [total - a for a in ans]

for a in ans:
    print(a % mod)