Problem:
Score : 200 points

Problem Statement

Takahashi has A cookies, and Aoki has B cookies.
Takahashi will do the following action K times:

If Takahashi has one or more cookies, eat one of his cookies.

Otherwise, if Aoki has one or more cookies, eat one of Aoki's cookies.

If they both have no cookies, do nothing.

In the end, how many cookies will Takahashi and Aoki have, respectively?

Constraints

0 \leq A \leq 10^{12}

0 \leq B \leq 10^{12}

0 \leq K \leq 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B K

Output

Print the numbers of Takahashi's and Aoki's cookies after K actions.

Sample Input 1

2 3 3

Sample Output 1

0 2

Takahashi will do the following:

He has two cookies, so he eats one of them.

Now he has one cookie left, and he eats it.

Now he has no cookies left, but Aoki has three, so Takahashi eats one of them.

Thus, in the end, Takahashi will have 0 cookies, and Aoki will have 2.

Sample Input 2

500000000000 500000000000 1000000000000

Sample Output 2

0 0

Watch out for overflows.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

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

/**
 * 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);
        B solver = new B();
        solver.solve(1, in, out);
        out.close();
    }

    static class B {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            // 整数の入力
            long A = Long.parseLong(in.next());
            long B = Long.parseLong(in.next());
            long K = Long.parseLong(in.next());
            long K2 = K - A;
            String ans = "";

            if (A >= K) {
                ans = String.valueOf(A - K) + " " + String.valueOf(B);
                out.println(ans);
                return;
            } else {
                ans = String.valueOf(0) + " " + String.valueOf(B - (K - A));
                out.println(ans);
                return;
            }
//        if(A>=B){
//            ans = String.valueOf(0) + " " + String.valueOf(K - B - A);
//            out.println(ans);
//            return;
//        }
//        if (A + B >= K) {
//            ans = String.valueOf(A-(K-B)) + " " + String.valueOf(B -(K-B));
//            out.println(ans);
//            return;
//        }
//        for (int i = 0; i < K; i++) {
//            if (A >= 1) {
//                A--;
//            }else if (B >= 1){
//                B--;
//
//            }
//        }
//
//        ans = String.valueOf(A) + " " + String.valueOf(B);
//        // 出力
//        out.println(ans);

        }

    }
}