import math

# Extended Euclidean Algorithm
def extgcd(a, b):
    if b:
        d, y, x = extgcd(b, a % b)
        y -= (a // b)*x
        return d, x, y
    return a, 1, 0

# lcm (least common multiple)
def lcm(m, n):
    return math.lcm(m,n)
def Chineese_mod_theory(b, m):
    # 求めたいxがあって m1とm2が互いに疎であり、x%m1≡b1, x%m2≡b2であるとき  
    # x = b1m2 + b2m1 である
    # 例えば b1m2%m1≡b1, b2m1%m1≡0である
    # これは拡張ユークリッドの互除法で考えると 
    # m1p1 + m2p2 = d(=gcd(m1, m2))
    # s = (b1-b2)/d　と置くと
    # m1p1s + m2p2s = b1-b2
    # x = b2 + m2p2s,  x = b1 - m1p1s
    # のどっちかを解けばいい
    MOD = 10**9 + 7
    M = 1
    r = 0
    for i in range(len(b)):
        d, p1, p2 = extgcd(M, m[i])
        if (b[i]-r)%d != 0:
            return (-1,0)
        s = (b[i]-r)//d
        tmp = (s * p1) % (m[i]//d)

        r = (r + M*tmp)
        M = (M * (m[i]//d))
    return (r%M)%MOD, M%MOD



def oi(): return int(input())
def os(): return input()
def mi(): return list(map(int, input().split()))

# import sys
# input = sys.stdin.readline
# import sys
# sys.setrecursionlimit(10**8)
# import pypyjit
# pypyjit.set_param('max_unroll_recursion=-1')
# input_count = 0

input_count = 0
N = oi()
amari = []
wari = []
all_zero = True
for _ in range(N):
    x,y = mi()
    wari.append(y)
    amari.append(x)
    if x != 0:
        all_zero = False
if all_zero:
    # amari, wari = pre_garner(amari, wari)
    out = 1
    for w in wari:
        out = (out*w)%(10**9+7)
    print(out)
else:
    # amari, wari = pre_garner(amari, wari)
    # if amari is None:
        # print(-1)
    # else:
    out = Chineese_mod_theory(amari, wari)
    print(out[0])