#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int main() {
    int N, K;
    cin >> N >> K;
    vector<int> A(N);
    for (int i = 0; i < N; ++i) {
        cin >> A[i];
    }

    // Separate the numbers into even and odd
    vector<int> even, odd;
    for (int num : A) {
        if (num % 2 == 0) {
            even.push_back(num);
        } else {
            odd.push_back(num);
        }
    }

    // Check if it's possible to form a subsequence of K elements with an even sum
    if (K == 0) {
        cout << "Yes" << endl;
        return 0; // Empty subsequence
    }

    // We need to have an even sum
    // Case 1: All numbers in subsequence are even
    if (even.size() >= K) {
        cout << "Yes" << endl;
        for (int i = 0; i < K; ++i) {
            cout << even[i] << " ";
        }
        cout << endl;
        return 0;
    }

    // Case 2: Use odd numbers
    int oddCount = odd.size();
    // We need an even count of odd numbers to keep the total sum even
    // (0, 2, 4, ...) odd numbers can be used
    int requiredOdds = (K - even.size()) % 2; // This should be even to maintain the parity
    
    // Check if we can satisfy the count
    if (even.size() + oddCount >= K && oddCount >= requiredOdds) {
        cout << "Yes" << endl;
        for (int i = 0; i < even.size(); ++i) {
            cout << even[i] << " ";
        }
        
        // Need to add odd numbers now
        int oddsToTake = K - even.size();
        if (oddsToTake % 2 != 0) {
            oddsToTake--; // Make it even
        }
        for (int i = 0; i < oddsToTake; ++i) {
            cout << odd[i] << " ";
        }
        cout << endl;
        return 0;
    }

    cout << "No" << endl;
    return 0;
}