1) 1234,2341,3412,4123
    2) 1243,2431,4312,3124
    3) 1324,3241,2413,4132
    4) 1342,3421,4213,2134
    5) 1423,4231,2314,3142
    6) 1432,4321,3214,2143

def beadOrnaments(b):
    sumColor=0
    intercolor=1
    for color in b:
        intercolor*=color**(color-1)
        sumColor+=color
    intercolor*=sumColor**(len(b)-2)
    return int(intercolor%(10**9+7))

private static final long MOD = 1000000007;

public static long power(long a, int pow) {
    long res = a % MOD;
    int count = 1;
    
    while (count < pow) {
        res = (res * a) % MOD;
        count++;
    }
    if (pow == 0) {
        return 1;
    }
    
    return res;
}

public static int beadOrnaments(List<Integer> b) {
// Write your code here
    int n = b.size();
    
    long sum = 0;
    long trees = 1;
    
    for (Integer i : b) {
        sum = (sum + i) % MOD;
        
        //this i - 2 + 1 is also from Cayley formula
        //first: number of sub trees are found using power(i, i - 2)
        //one leaf of a sub tree will connect with root of another
        //   sub tree
        //hence there are i ways of connecting as there should be 
        //   i leaves in each tree
        //therefore power(i, i - 2) * i == power(i, i - 2 + 1)
        
        trees = (trees * power(i, i - 2 + 1)) % MOD;
    }
    
    if (n >= 2) {
        
        //Cayley formular 
        //https://www.geeksforgeeks.org/cayleys-formula/
        
        trees = (trees * power(sum, n - 2)) % MOD;
    }
    else {
        trees = (long) Math.floor(trees / sum);
    }
            
    return (int) (trees % MOD); 
}

import  functools

def beadOrnaments(b):
    ans = functools.reduce(lambda ans, a: ans * (a ** (a - 1)), b,1)
    sm = functools.reduce(lambda sm, a: sm + a, b,0)
    if (len(b) - 2 >= 0):
        ans *= sm ** (len(b) - 2)
    else:
        ans //= sm
    return ans % 1000000007