Vasya has got three integers n, m and k. He'd like to find three integer points (x_1, y_1), (x_2, y_2), (x_3, y_3), such that 0 ≤ x_1, x_2, x_3 ≤ n, 0 ≤ y_1, y_2, y_3 ≤ m and the area of the triangle formed by these points is equal to nm/k.

Help Vasya! Find such points (if it's possible). If there are multiple solutions, print any of them.

Input

The single line contains three integers n, m, k (1≤ n, m ≤ 10^9, 2 ≤ k ≤ 10^9).

Output

If there are no such points, print "NO".

Otherwise print "YES" in the first line. The next three lines should contain integers x_i, y_i — coordinates of the points, one point per line. If there are multiple solutions, print any of them.

You can print each letter in any case (upper or lower).

Examples

Input

4 3 3


Output

YES
1 0
2 3
4 1


Input

4 4 7


Output

NO

Note

In the first example area of the triangle should be equal to nm/k = 4. The triangle mentioned in the output is pictured below: 

<image>

In the second example there is no triangle with area nm/k = 16/7.
import java.io.*; 
import java.math.BigInteger;
import java.util.*;

public class Main {
	static final long MOD = 998244353;
    public static void main(String[] args) throws IOException {
        FastScanner sc=new FastScanner();
        long N = sc.nextLong();
        long M = sc.nextLong();
        long[] bounds = {N,M};
        long K = sc.nextLong();
        long NKgcd = gcd(N,K);
        N /= NKgcd;
        K /= NKgcd;
        long MKgcd = gcd(M,K);
        M /= MKgcd;
        K /= MKgcd;
        if (K > 2) {
        	System.out.println("NO");
        	return;
        } else if (K == 1) {
        	if (2*N <= bounds[0]) {
        		N *= 2;
        		K = 2;
        	} else {
        		M *= 2;
        		K = 2;
        	}
        }
        
        System.out.println("YES");
        System.out.println(0 + " " + 0);
        System.out.println(N + " " + 0);
        System.out.println(0 + " " + M);
    }
    
    public static long gcd(long p, long q) {
    	if (p < q) {
    		return gcd(q,p);
    	}
        while (q != 0) {
            long temp = q;
            q = p % q;
            p = temp;
        }
        return p;
    }
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() 
        { 
            br = new BufferedReader(new
                     InputStreamReader(System.in)); 
        } 
  
        String next() 
        { 
            while (st == null || !st.hasMoreElements()) 
            { 
                try
                { 
                    st = new StringTokenizer(br.readLine()); 
                } 
                catch (IOException  e) 
                { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() 
        { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() 
        { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() 
        { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() 
        { 
            String str = ""; 
            try
            { 
                str = br.readLine(); 
            } 
            catch (IOException e) 
            { 
                e.printStackTrace(); 
            } 
            return str; 
        } 
    }
}

class Node {
	public int n;
	public HashSet<Node> children;
	
	public Node(int n) {
		this.n = n;
		children = new HashSet<Node>();
	}
	
	public void addChild(Node node) {
		children.add(node);
	}
}

class BinaryIndexedTree {
	  public long[] arr;

	  public BinaryIndexedTree (int N) {
	    arr = new long[N+1];
	    arr[0] = 0;
	  }

	  //add k to the i-th element.
	  public void add(long k, int i) {
	    int node = i+1;
	    while (node < arr.length) {
	      arr[node] += k;
	      node += node & (-node);
	    }
	  }

	  //sum up the elements from input[s_i] to input[e_i], from [s_i,e_i).
	  public long sum(int s_i, int e_i) {
	    return sum(e_i) - sum(s_i);
	  }

	  public long sum(int i) {
	    long total = 0;
	    int node = i;
	    while (node > 0) {
	      total += arr[node];
	      node -= node & (-node);
	    }
	    return total;
	  }
}