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.util.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }
}

class TaskD {
    long MOD = (long)1e9 + 7;
    long p[];

    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int n = in.nextInt();
        int m = in.nextInt();
        int g = in.nextInt();
        p = new long[m + n + 1];
        p[0] = 1;
        for(int i = 1; i < p.length; ++i) p[i] = p[i - 1] * i % MOD;

        long res = 0;
        if(g == 1)
            res = rec(n, m);
        else if(g == 0) {
            if(n == 0 && m == 0)
                res = 0;
            else if(n == 1 && m == 0)
                res = 1;
            else if(n == 0 && m == 1)
                res = 0;
            else {
                if(n >= 1)
                    res = (res + rec(n - 1, m)) % MOD;
                if(m >= 1)
                    res = (res + p[n + m - 1] * modPow(p[n] * p[m - 1] % MOD, MOD - 2) % MOD) % MOD;
            }
        }
        out.println(res);
    }

    private long rec(int n, int m) {
        long res = 0;
        for(int i = n; i >= 0; ) {
            if(i == 1) {
                if(m <= 1) return res;
                else if(m > 1) {res = (res + 1) % MOD; return res;}
            }
            else if(i == 0) {
                if(m == 0 || m > 1) return res;
                else if(m == 1) {res = (res + 1) % MOD; return res;}
            }
            else if(i == 2 && m == 0) {res = (res + 1) % MOD; return res;}
            else if(i == 0 && m == 2) return res;

            else {
                if(i > 1 && m >= 1)
                    res = (res + p[m - 1 + i - 1] * modPow(p[m - 1] * p[i - 1] % MOD, MOD - 2) % MOD) % MOD;
                i -= 2;
            }
        }
        return res;
    }

    private long modPow(long a, long p) {
        if(p == 0) return 1;
        long res = modPow(a, p / 2);
        res = res * res % MOD;
        if(p % 2 == 1)
            res = res * a % MOD;
        return res;
    }
}
