"""

それぞれの数について

(3* 2^0 + 2* 2^1 + 1* 2^2) = 3+4+4 = 11
(2* 3^0 + 1* 3^1) = 2+3 = 5

みたいにすればよい

Σ()

"""

import sys
from sys import stdin

mod = 10**9+7

def inv(x,mod):
    return pow(x,mod-2,mod)

def solve(e,P):

    N = e+1

    ret = P * ( pow(P,N,mod)-1 ) + N*(-P) + N
    ret *= inv( (P-1)**2 , mod )
    return ret % mod

L = int(stdin.readline())

ans = 1

for i in range(L):

    P,e = map(int,stdin.readline().split())
    ans *= solve(e,P)
    ans %= mod

print (ans % mod)