Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const int MAXN = 2e5 + 10;
const int MAXM = 1e4 + 500;
const int N = 5200 + 15;
const double EPS = 1e-9;
int dx[] = {1, 0, 0, -1};
int dy[] = {0, 1, -1, 0};
double len(complex<long long> a) {
  return sqrt(a.real() * a.real() + a.imag() * a.imag());
}
long long sq(complex<long long> a) {
  return a.real() * a.real() + a.imag() * a.imag();
}
complex<long long> perp(complex<long long> p) { return {-p.imag(), p.real()}; }
long long dotProduct(complex<long long> a, complex<long long> b) {
  return (conj(a) * b).real();
}
long long crossProduct(complex<long long> a, complex<long long> b) {
  return (conj(a) * b).imag();
}
long long orient(complex<long long> a, complex<long long> b,
                 complex<long long> c) {
  return crossProduct(b - a, c - a);
}
bool inDisk(complex<long long> a, complex<long long> b, complex<long long> p) {
  return dotProduct(a - p, b - p) <= 0;
}
bool onSegment(complex<long long> a, complex<long long> b,
               complex<long long> p) {
  return orient(a, b, p) == 0 && inDisk(a, b, p);
};
struct cmpX {
  bool operator()(complex<long long> a, complex<long long> b) {
    return make_pair(a.real(), a.imag()) < make_pair(b.real(), b.imag());
  }
};
struct line {
  complex<long long> v;
  long long c;
  line(complex<long long> v, long long c) : v(v), c(c) {}
  line(long long a, long long b, long long c) : v({b, -a}), c(c) {}
  line(complex<long long> p, complex<long long> q)
      : v(q - p), c(crossProduct(v, p)) {}
  long long side(complex<long long> p) { return crossProduct(v, p) - c; }
  double dist(complex<long long> p) { return abs(side(p)) / len(v); }
  line perpThrough(complex<long long> p) { return {p, p + perp(v)}; }
  bool cmpProj(complex<long long> p, complex<long long> q) {
    return dotProduct(v, p) < dotProduct(v, q);
  }
  line translate(complex<long long> t) { return {v, c + crossProduct(v, t)}; }
  complex<long long> proj(complex<long long> p) {
    return p - perp(v) * side(p) / sq(v);
  }
};
double segPoint(complex<long long> a, complex<long long> b,
                complex<long long> p) {
  if (a != b) {
    line l(a, b);
    if (l.cmpProj(a, p) && l.cmpProj(p, b)) return l.dist(p);
  }
  return min(abs(p - a), abs(p - b));
}
bool inAngle(complex<long long> a, complex<long long> b, complex<long long> c,
             complex<long long> p) {
  if (orient(a, b, c) < 0) swap(b, c);
  return orient(a, b, p) >= 0 && orient(a, c, p) <= 0;
}
bool isInsidePol(vector<complex<long long> > &a, complex<long long> b) {
  int n = a.size();
  int pos = 0, neg = 0;
  for (int i = 0; i < n - 1; i++) {
    if (onSegment(a[i], a[(i + 1) % n], b)) return false;
    line l(a[i], a[i + 1]);
    if (l.side(b) > 0)
      pos++;
    else
      neg++;
  }
  line l(a[n - 1], a[0]);
  if (l.side(b) > 0)
    pos++;
  else
    neg++;
  return ((!pos && neg == n) || (!neg && pos == n));
}
long long i, j, k;
long long n, m, t;
long long a[MAXN];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int p, k;
  cin >> n >> k >> p;
  vector<int> evens, odds;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    if (a[i] % 2)
      odds.push_back(a[i]);
    else
      evens.push_back(a[i]);
  }
  int q = k - p;
  if (odds.size() < q) {
    cout << "NO";
    return 0;
  }
  int remOdd = odds.size() - q;
  if (remOdd / 2 + evens.size() < p || remOdd % 2) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  if (p > 0) {
    for (i = 0; i < q; i++) cout << 1 << ' ' << odds[i] << endl;
    copy(odds.begin() + q, odds.end(), back_inserter(evens));
    for (i = j = 0; j < p - 1; j++, i++) {
      if (evens[i] & 1) {
        cout << 2 << ' ' << evens[i] << ' ' << evens[i + 1] << endl;
        i++;
      } else {
        cout << 1 << ' ' << evens[i] << endl;
      }
    }
    cout << evens.size() - i << ' ';
    for (; i < evens.size(); i++) cout << evens[i] << ' ';
  } else {
    copy(evens.begin(), evens.end(), back_inserter(odds));
    for (i = 0; i < q - 1; i++) cout << 1 << ' ' << odds[i] << endl;
    cout << odds.size() - (q - 1) << ' ';
    for (; i < odds.size(); i++) cout << odds[i] << ' ';
  }
  return 0;
}