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.
#http://codeforces.com/problemset/problem/615/D
def binaryExpo(a,x,modulo):
    if x == 0:
        return 1
    y = binaryExpo(a,x/2,modulo)
    if x%2 == 0:
        return (y*y)%modulo
    else:
        return (a*y*y)%modulo

def fun0(primes,primePowers,modulo):
    constantPower = 1
    for alpha in primePowers:
        constantPower = (constantPower*(alpha+1))%(modulo-1)
    i = 0
    numOfPrimes = len(primes)
    productOfDivisors = 1
    while i < numOfPrimes:
        power = ((primePowers[i]/2)*(constantPower))%(modulo-1)
        productOfDivisors = (productOfDivisors*binaryExpo(primes[i],power,modulo))%modulo
        i += 1
    return productOfDivisors

def fun1(primes,primePowers,modulo,flaggedAlpha):
    constantPower = 1
    encountered = 0
    for alpha in primePowers:
        if alpha == flaggedAlpha and encountered == 0:
            x = (alpha+1)/2
            constantPower = (constantPower*x)%(modulo-1)
            encountered = 1
        else:
            constantPower = (constantPower*(alpha+1))%(modulo-1)
    i = 0
    numOfPrimes = len(primes)
    productOfDivisors = 1
    while i < numOfPrimes:
        power = ((primePowers[i])*(constantPower))%(modulo-1)
        productOfDivisors = (productOfDivisors*binaryExpo(primes[i],power,modulo))%modulo
        i += 1
    return productOfDivisors


m = int(raw_input())
A = map(int,raw_input().split())
A.sort()
prev = A[0]
primeCounter = 1
primes = []
primePowers = []
i = 1
while i < m:
    if A[i] == prev:
        primeCounter += 1
    else:
        primes.append(prev)
        primePowers.append(primeCounter)
        primeCounter = 1
        prev = A[i]
    i += 1
primes.append(A[m-1])
primePowers.append(primeCounter)

modulo = 10**9+7
#flag = 0 corresponds to where all the prime powers are even
flag  = 0
for alpha in primePowers:
    if (alpha+1)%2 == 0:
        flaggedAlpha = alpha
        flag = 1
        break
if flag == 0:
    print fun0(primes,primePowers,modulo)
else:
    print fun1(primes,primePowers,modulo,flaggedAlpha)

    
