Ayrat has number n, represented as it's prime factorization pi of size m, i.e. n = p1·p2·...·pm. Ayrat got secret information that that the product of all divisors of n taken modulo 109 + 7 is the password to the secret data base. Now he wants to calculate this value.

Input

The first line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of primes in factorization of n. 

The second line contains m primes numbers pi (2 ≤ pi ≤ 200 000).

Output

Print one integer — the product of all divisors of n modulo 109 + 7.

Examples

Input

2
2 3


Output

36


Input

3
2 3 2


Output

1728

Note

In the first sample n = 2·3 = 6. The divisors of 6 are 1, 2, 3 and 6, their product is equal to 1·2·3·6 = 36.

In the second sample 2·3·2 = 12. The divisors of 12 are 1, 2, 3, 4, 6 and 12. 1·2·3·4·6·12 = 1728.
mod=10**9+7

def pow(a,b): #Algoritmo de Exponenciacion binaria modular
    exp=1 # Caso base a^1=a
    x=a % mod
    b=b%(mod-1) #
    while b > 0: 
        if b % 2 == 1:# En caso que b sea impar
            exp=(exp*x)%mod
            # En caso que b sea impar
        x=(x*x)%mod
        b=b//2
        b=b%(mod-1)
    return exp%mod   

x=input() #
primes=map(int,input().split())# Capturo la entrada del problema
primesDicc=dict() # inicializo un diccionario para guardar la descomposicion en primos de la forma <pi,e1>donde e1 es el exponente con que aparece pi en la descomposicion
n=1;
for p in primes: 
    n=(n*p)%mod
    if p not in primesDicc: 
        primesDicc[p]=1
    else:
        primesDicc[p]+=1
 
# Mapping(primes)
countdiv =1
raiz=1
for p in primesDicc: #Hago un recorrido por todos las llaves de mi diccionario para generar el numero de divisores
    countdiv *= primesDicc[p]+1 
result=0
if countdiv % 2 == 0:
    result=pow(n,countdiv//2)
else:
    for p in primesDicc:
        raiz=(raiz*pow(p,primesDicc[p]//2))
    result=(raiz*pow(n,countdiv//2))%mod  

print(result)