import java.util.*;
import java.io.*; 
import java.text.*;
//Solution Credits: Taranpreet Singh
class Editorial{
    //SOLUTION BEGIN
    int B = 32;
    long[][][][] dp = new long[B][B][B][];
    void solve(int TC) throws Exception{
        //CLeared DP array
        for(int i = 0; i< B; i++)
            for(int j = 0; j< B; j++)
                for(int k =0; k< B; k++)
                    dp[i][j][k] = new long[]{-1,-1};
        long a = nl(), b = nl(), c = nl();
        pn(ans(0,bit(a), bit(b),0, c));
    }
    //bit = bit :P, a = Number of bits of A yet to be used, b = Number of bits of B yet to be used, cf = carry forward, c is same as given in input.
    long ans(int bit, int a, int b, int cf, long c){
        //Impossible case, Number of bits available cannot be negative
        if(a<0 || b<0)return 0;
        //Base case, Reached 32nd bit.
        if(bit==B)return (a==0 && b==0 && cf==0)?1:0;
        //If answer for this state already calculated.
        if(dp[bit][a][b][cf]!=-1)return dp[bit][a][b][cf];
        //Current bit of c is 0 or 1
        int x = (c&(1l<<bit))>0?1:0;
        long ans = 0;
        //If cf is same as required bit, either bit is placed in both numbers, or none of the number.
        //If bit isn't used in any number, carry is 0 always. If bit is used in both numbers, carry is 1.
        if(cf==x)ans+=ans(bit+1,a,b,0,c)+ans(bit+1,a-1,b-1,1,c);
        //If cf is different, we need one of either A or B to have current bit 1, and other one should have bit 0.
        //Carry forward is 1, if cf is 1, otherwise 0.
        if(cf%2!=x)ans+=ans(bit+1, a-1,b,(cf+1)/2, c)+ans(bit+1,a,b-1,(cf+1)/2, c);
        return dp[bit][a][b][cf] = ans;
    }
    //SOLUTION ENDS
    long mod = (long)1e9+7, IINF = (long)1e17;
    final int MAX = (int)1e6+1, INF = (int)2e9, root = 3;
    DecimalFormat df = new DecimalFormat("0.000000000000");
    double PI = 3.1415926535897932384626433832792884197169399375105820974944, eps = 1e-8;
    static boolean multipleTC = true, memory = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        in = new FastReader();
        out = new PrintWriter(System.out);
        int T = (multipleTC)?ni():1;
        //Solution Credits: Taranpreet Singh
        for(int i = 1; i<= T; i++)solve(i);
        out.flush();
        out.close();
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Editorial().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Editorial().run();
    }
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n(){return in.next();}
    String nln(){return in.nextLine();}
    int ni(){return Integer.parseInt(in.next());}
    long nl(){return Long.parseLong(in.next());}
    double nd(){return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next(){
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        String nextLine(){
            String str = "";
            try{    
                str = br.readLine();
            }catch (IOException e){
                e.printStackTrace();
            }   
            return str;
        }
    }
}