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.
The first line will contain three space separated integers *n*, *k*, *p* (1 ≤ *k* ≤ *n* ≤ 10<sup class="upper-index">5</sup>; 0 ≤ *p* ≤ *k*). The next line will contain *n* space-separated distinct integers representing the content of array *a*: *a*<sub class="lower-index">1</sub>, *a*<sub class="lower-index">2</sub>, ..., *a*<sub class="lower-index">*n*</sub> (1 ≤ *a*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>).
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace __gnu_pbds;  
using namespace std;
 
#define int                 long long int
#define ull                 unsigned int
#define ld                  long double
#define pb                  push_back
#define ff                  first
#define ss                  second
#define S(arr)              sort(arr.begin(),arr.end());
#define RS(arr)              sort(arr.rbegin(),arr.rend());
#define R(arr)              reverse(arr.begin(),arr.end());
#define si                  set <int>
#define vi                  vector <int>
#define setbits(x)          __builtin_popcountll(x)
#define zrobits(x)          __builtin_ctzll(x)
#define vvi                 vector <vector <int>>
#define vs                  vector <string>
#define pii                 pair <int, int>
#define vpi                 vector <pii>
#define vpp                 vector <pair<int, pii>>
#define mii                 map <int, int>
#define mci                 map<char, int>
#define mpi                 map <pii, int>
#define spi                 set <pii>
#define speed               ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define endl                "\n"
#define all(p)              p.begin(), p.end()
#define T                   int tcf;cin>>tcf;while(tcf--)
#define que_max(t)          priority_queue <t>
#define que_min(t)          priority_queue <t, vector<t>, greater<t>>
#define dba(a, l, r)        for(auto x:a)cout<<x<<" ";cout<<endl;
#define db1(x)              cout << #x " = " << x << endl;
#define db2(x,y)            cout << #x " = " << x << " " << #y " = " << y << endl;
#define db3(x,y,z)          cout << #x " = " << x << " " << #y " = " << y << " " << #z " = " << z << endl;
#define db4(x,y,z,w)        cout << #x " = " << x << " " << #y " = " << y << " " << #z " = " << z << " " << #w " = " << w << endl;
#define db5(x,y,z,w,v)      cout << #x " = " << x << " " << #y " = " << y << " " << #z " = " << z << " " << #w " = " << w << " " << #v " = " << v << endl;
#define db6(x,y,z,w,v,a) db3(a, y, x)db3(w, v, a)
#define p2d(v)              for(auto a:v){for(auto b:a)cout<<b<<" ";cout<<endl;}
#define p1d(v)              for(auto a:v)cout<<a<<" ";cout<<endl;
#define ppi(v)              for(auto a:v)cout<<a.ff<<" "<<a.ss<<endl;


typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
 
void read(int n=10){
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
    cout<<setprecision(n)<<fixed;
    #ifndef ONLINE_JUDGE
         freopen("input.txt",  "r",  stdin);
         freopen("output.txt", "w", stdout);
    #endif
}
 
int M = 1e9 + 7;
int M1 = 998244353;
#define qqq 4000005
 
 
int32_t main(){
    read();
    int n ,k,p;
    cin>>n>>k>>p;
    vi arr(n);
    vi odd,even;
    for(int i=0;i<n;i++){
        cin>>arr[i];
        if(arr[i]%2==0){
            even.pb(arr[i]);
        }else{
            odd.pb(arr[i]);
        }
    }
    int o=odd.size();
    int e = n-o;
    int lo = o-(k-p);
    int io=0,ie=0;
    // db6(o,e,lo,k,p,n)
    if(k-p<=o){
        if(p<=e+(lo)/2&&lo%2==0){
            cout<<"YES\n";
            for(int i=0;i<k-p;i++){
                cout<<1<<" "<<odd[io++]<<endl;
            }for(int i=0;i<p-1;i++){
                cout<<1<<" "<<even[ie++]<<endl;
            }
            vi res;
            while(io<odd.size())
                res.pb(odd[io++]);
            while(ie<even.size())
                res.pb(even[ie++]);
            cout<<res.size()<<" ";
            p1d(res);
            return 0; 
        }
    }
    cout<<"NO";
}  