Vasily the Bear loves beautiful strings. String s is beautiful if it meets the following criteria: 

  1. String s only consists of characters 0 and 1, at that character 0 must occur in string s exactly n times, and character 1 must occur exactly m times. 
  2. We can obtain character g from string s with some (possibly, zero) number of modifications. The character g equals either zero or one. 



A modification of string with length at least two is the following operation: we replace two last characters from the string by exactly one other character. This character equals one if it replaces two zeros, otherwise it equals zero. For example, one modification transforms string "01010" into string "0100", two modifications transform it to "011". It is forbidden to modify a string with length less than two.

Help the Bear, count the number of beautiful strings. As the number of beautiful strings can be rather large, print the remainder after dividing the number by 1000000007 (109 + 7). 

Input

The first line of the input contains three space-separated integers n, m, g (0 ≤ n, m ≤ 105, n + m ≥ 1, 0 ≤ g ≤ 1).

Output

Print a single integer — the answer to the problem modulo 1000000007 (109 + 7).

Examples

Input

1 1 0


Output

2


Input

2 2 0


Output

4


Input

1 1 1


Output

0

Note

In the first sample the beautiful strings are: "01", "10".

In the second sample the beautiful strings are: "0011", "1001", "1010", "1100".

In the third sample there are no beautiful strings.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;


public class D {	
	BufferedReader reader;
    StringTokenizer tokenizer;
    PrintWriter out;
    
    long MOD = 1000000007;
    public long mod_inverse(long n1){
    	return mod_inverse(n1, MOD);
    }
    public long mod_inverse(long x, long p) {
    	// x^(-1) mod p
		return BigInteger.valueOf(x).modInverse(BigInteger.valueOf(p)).longValue();
	}
    public long mod_add(long n1, long n2){
    	return (n1 + n2) % MOD;
    }    
    public long mod_time(long n1, long n2){
    	return (n1 * n2) % MOD;
    }
    
	public void solve() throws IOException {
		int M = nextInt();
		int N = nextInt();
		
		int G = nextInt();
	
		if( N == 0){
			if( G == 1){
				if( M % 2 == 0) out.println(1);
				else out.println(0);
			}
			else{
				if( M % 2 != 0) out.println(1);
				else out.println(0);
			}
			return;
		}
		
		long ans = 0;
		
		long rollPoll = 1;
			
		for(int i = M; i >= 0; i--){					
			if(G == 1){
				if(N == 1 && i == M){
					if( i % 2 == 0) ans = mod_add(ans, 1);
				}
				else{
					if( i % 2 == 1) ans = mod_add(ans, rollPoll );
				}
			}
			else{
				if(N == 1 && i == M){
					if( i % 2 != 0) ans = mod_add(ans, 1);
				}
				else{
					if( i % 2 == 0) ans = mod_add(ans, rollPoll );
				}
			}
			
			rollPoll = mod_time(rollPoll, N + M - i);
			rollPoll = mod_time(rollPoll, mod_inverse(M-i+1));
			
		}
		out.println( ans );
	}
	
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		new D().run();
	}
	
	public void run() {
        try {
            reader = new BufferedReader(new InputStreamReader(System.in));
            tokenizer = null;
            out = new PrintWriter(System.out);
            solve();
            reader.close();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }

    String nextToken() throws IOException {
        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

}
