#yukicoder 187 中華風(Hard)  多倍長解法

#exec gcd
#Reference: https://x.com/_miz_tom/status/1842086118990479854
exec('def gcd(x, y):\n' +
     ' x, y = abs(x), abs(y)\n' +
     ' if x == 0: return y\n' +
     ' if not (y := y % x): return x\n if not (x := x % y): return y\n' * 300 +
     ' return gcd(x, y)')
     
#exec gcdを使わない版
gcd = lambda x, y: gcd(y, x % y) if y else abs(x)
def CRT(m1, M1, m2, M2):  #smallest n ≡ m1 mod M1 ≡ m2 mod M2
    if (m1 - m2) % ( G := gcd(M1, M2) ) != 0: return -1
    return (m2 - m1) * pow(M1 // G, -1, M2 // G) % M2 * M1 // G + m1


#入力受取
N = int(input())
P = [tuple(map(int, input().split())) for _ in range(N)]
MOD = 10 ** 9 + 7

#CRTを実行
rem, mod = 0, 1
for Xi, Yi in P:
    rem = CRT(rem, mod, Xi, Yi)
    if rem == -1:
        break
    mod = mod * Yi // gcd(mod, Yi)

if   rem == -1:  #解なし
    print(-1)
elif rem == 0:  #正整数を探す → lcm(Yi)
    print(mod % MOD)
else:  #remを出力
    print(rem % MOD)