Codeforces Round 251 (Div. 2)
C
Devu and Partitioning of the Array
Codeforces
1700
原题
← BD →
翻译中…

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?

翻译中…

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).

翻译中…

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

翻译中…

Input

翻译中…

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 $*a*$: $*a*_{1}, *a*_{2}, ..., *a*_{*n*}$ ($1 ≤ *a*_{*i*} ≤ 10^{9}$).

翻译中…

Output

翻译中…

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).

翻译中…

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.

翻译中…

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

翻译中…

样例

Input #1
5 5 3
2 6 10 5 9
Output #1
YES
1 9
1 5
1 10
1 6
1 2
InputCopy OutputCopy
Input #2
5 5 3
7 14 2 9 5
Output #2
NO
InputCopy OutputCopy
Input #3
5 3 1
1 2 3 7 5
Output #3
YES
3 5 1 3
1 7
1 2
InputCopy OutputCopy
题目信息
题号
439C
时间限制
1 s
内存限制
256 MB
标签
暴力
构造
模拟
数论

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