Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?

Given an array consisting of distinct integers. Is it possible to partition the whole array into k disjoint non-empty parts such that p of the parts have even sum (each of them must have even sum) and remaining k - p have odd sum? (note that parts need not to be continuous).

If it is possible to partition the array, also give any possible way of valid partitioning.

Input

The first line will contain three space separated integers n, k, p (1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k). The next line will contain n space-separated distinct integers representing the content of array a: a1, a2, ..., an (1 ≤ ai ≤ 109).

Output

In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).

If the required partition exists, print k lines after the first line. The ith of them should contain the content of the ith part. Print the content of the part in the line in the following way: firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. There must be exactly p parts with even sum, each of the remaining k - p parts must have odd sum.

As there can be multiple partitions, you are allowed to print any valid partition.

Examples

Input

5 5 3
2 6 10 5 9


Output

YES
1 9
1 5
1 10
1 6
1 2


Input

5 5 3
7 14 2 9 5


Output

NO


Input

5 3 1
1 2 3 7 5


Output

YES
3 5 1 3
1 7
1 2
"""
Devu being a small kid, likes to play a lot, but he only likes to play with arrays. While playing he came up with an interesting question which he could not solve, can you please solve it for him?

Given an array consisting of distinct integers. Is it possible to partition the whole array into 
k disjoint non-empty parts such that 
p of the parts have EVEN sum (each of them must have even sum) and remaining 
k - p have ODD sum? 
(note that parts need not to be continuous).

If it is possible to partition the array, also give any possible way of valid partitioning.

Input
The first line will contain three space separated integers n, k, p (1 ≤ k ≤ n ≤ 105; 0 ≤ p ≤ k). The next line will contain n space-separated distinct integers representing the content of array a: a1, a2, ..., an (1 ≤ ai ≤ 109).

Output
In the first line print "YES" (without the quotes) if it is possible to partition the array in the required way. Otherwise print "NO" (without the quotes).

If the required partition exists, print k lines after the first line. The ith of them should contain the content of the ith part. Print the content of the part in the line in the following way: 
firstly print the number of elements of the part, then print all the elements of the part in arbitrary order. 
There must be exactly p parts with even sum, each of the remaining k - p parts must have odd sum.

As there can be multiple partitions, you are allowed to print any valid partition.

input
5 5 3
2 6 10 5 9
output
YES
1 9
1 5
1 10
1 6
1 2

input
5 5 3
7 14 2 9 5
output
NO

input
5 3 1
1 2 3 7 5
output
YES
3 5 1 3
1 7
1 2

"""


class part_array:
    def __init__(self):
        self.count = 0
        self.num_list = []
    def AddNumber(self, n):
        self.num_list.append(n)
        self.count += 1
    def AddNumberList(self, li):
        self.num_list.extend(li)
        self.count += len(li)


def print_solution_and_exit(e, o):
    print('YES')
    for part in e:
        print (part.count, ' '.join(map(str, part.num_list)))
    for part in o:
        print (part.count, ' '.join(map(str, part.num_list)))
    exit()

def print_solution_error():
    print ('NO')
    exit()


n, k, p = map(lambda t : int(t), input().split())
numbers = list(map(lambda t : int(t), input().split()))

odds = list(filter( lambda t: t % 2 == 1, numbers))
evens = list(filter( lambda t: t % 2 == 0, numbers))

odd_parts = []
even_parts = []

# need to have at leas k - p odd numbers. Cannot get an odd sum without one
if k - p > len(odds):
    print_solution_error()

# check if we can get the even piles
#p - len(evens) < 


while len(even_parts) < p and len(evens) > 0:
    part = part_array()
    part.AddNumber(evens.pop())
    even_parts.append(part)

while len(odd_parts) < k - p and len(odds) > 0:
    part = part_array()
    part.AddNumber(odds.pop())
    odd_parts.append(part)

# check if only odd parts need to be created
needed_odd_parts = k - p - len(odd_parts)
if len(even_parts) == p and needed_odd_parts > 0:
    # excess even numbers can be added to any odd part
    excess_odds = len(odds) - needed_odd_parts
    if excess_odds < 0 or excess_odds % 2 == 1:
        # not enough odd numberss
        # the exxess need to have an even sum
        print_solution_error()

    for i in range(needed_odd_parts):
        part = part_array()
        part.AddNumber(odds.pop())
        odd_parts.append(part)

#check if only even parts need to be created
needed_even_parts = p - len(even_parts)
if len(odd_parts) == k - p and needed_even_parts > 0:
    excess_evens = len(evens) - needed_even_parts
    if len(odds) % 2 == 1 or excess_evens + len(odds) // 2 < 0:
        # not enough odd numberss
        # the exxess need to have an even sum
        print_solution_error()

    while  needed_even_parts > 0:
        if len(evens) > 0:
            part = part_array()
            part.AddNumber(evens.pop())
            even_parts.append(part)
        else:
            part = part_array()
            part.AddNumber(odds.pop())
            part.AddNumber(odds.pop())
            even_parts.append(part)
        needed_even_parts -= 1


# now we need both even and odd, but we exausted all low hanging
needed_even_parts = p - len(even_parts)
needed_odd_parts = k - p - len(odd_parts)
if needed_odd_parts > 0 or needed_even_parts > 0:
    print('needed_even_parts', needed_even_parts, 'needed_odd_parts', needed_odd_parts)
    print_solution_and_exit()

# here we just add the remaning numbers to the solution
if len(odds) % 2 == 1:
    print_solution_error()

#k is at least 1 so we have a part
if len(odd_parts) > 0:
    odd_parts[0].AddNumberList(odds)
    odd_parts[0].AddNumberList(evens)
else:
    even_parts[0].AddNumberList(odds)
    even_parts[0].AddNumberList(evens)


ods = [] # done
evens = [] #done

print_solution_and_exit(even_parts, odd_parts)
    


#while len(odd_parts) + len(even_parts) < k:




print('DEBUG')
print (numbers)
print (evens)
print (odds)