Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

C. Devu and Partitioning of the Array

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputDevu 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* ≤ 10^{5}; 0 ≤ *p* ≤ *k*). The next line will contain *n* space-separated distinct integers representing the content of array *a*: *a*_{1}, *a*_{2}, ..., *a*_{n} (1 ≤ *a*_{i} ≤ 10^{9}).

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 *i*^{th} of them should contain the content of the *i*^{th} 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

Codeforces (c) Copyright 2010-2017 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Dec/15/2017 13:00:13 (c5).

Desktop version, switch to mobile version.

User lists

Name |
---|