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.*;
import java.util.Scanner;

/**
 * @author pvasilyev
 * @since 8/11/13
 */
public class ProblemD {

    private static final long MOD = 1000 * 1000 * 1000 + 7;

    public static final String FILE_IN = "std.in";
    public static final String FILE_OUT = "std.out";
    private static boolean debugMode = true;
    private static long[] FACT = new long[200001];

    static {
        long m = 1;
        FACT[0] = 1;
        for (int k = 1; k < FACT.length; ++k) {
            m = (m * k) % MOD;
            FACT[k] = m;
        }
    }

    public static void main(String[] args) throws IOException {

        final Scanner reader = new Scanner((debugMode ? System.in : new FileInputStream(FILE_IN)));
        final PrintWriter writer = new PrintWriter(debugMode ? System.out : new FileOutputStream(FILE_OUT));

        solveTheProblem(reader, writer);

        reader.close();
        writer.close();
    }

    private static void solveTheProblem(final Scanner reader, final PrintWriter writer) throws IOException {
        final int n = reader.nextInt();
        final int m = reader.nextInt();
        final int g = reader.nextInt();

        long sum = 0;

        if (n == 0) {
            if (g == 0 && m > 1 || g == 1 && m == 1) {
                writer.println("1");
            } else {
                writer.println("0");
            }
        } else if (m == 0) {
            if (g == 1 && n % 2 == 0) {
                writer.println("1");
            } else if (g == 0 && n % 2 == 1) {
                writer.println("1");
            } else {
                writer.println("0");
            }
        } else {
            for (int k = 0; k <= n; k++) {
                final int n1 = n + m - k - 1;
                if (n1 > 0) {
                    if (k % 2 == g) {
                        sum = (sum + C(n1, m - 1)) % MOD;
                    }
                } else {
                    if (k % 2 != g) {
                        sum = (sum + C(n1, m - 1)) % MOD;
                    }
                }
            }
            writer.println(sum);
        }

    }

    private static long C(final int n, final int k) {
        final long temp = (FACT[n - k] * FACT[k]) % MOD;
        return (FACT[n] * inv(temp, MOD)) % MOD;
    }

    static long inv(long value, final long mod) {
        long pow = mod - 2;

        long exp = value;
        value = quickPow(exp, pow, mod);
        return value % mod;
    }

    private static long quickPow(final long exp, final long pow, final long mod) {
        if (pow == 0) {
            return 1;
        } else if (pow == 1) {
            return exp;
        } else if (pow % 2 == 0) {
            long tmp = quickPow(exp, pow / 2, mod);
            return (tmp * tmp) % mod;
        } else {
            long tmp = quickPow(exp, pow / 2, mod);
            tmp = (tmp * tmp) % mod;
            return (tmp * exp) % mod;
        }
    }

}
