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 C++14 (GCC 5.4.1) (judge verdict: Accepted)


#include <bits/stdc++.h>
#define PRECISION(x) cout << fixed << setprecision(x); 
#define FAST_IO ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define SZ(X) ((int)(X).size())
#define ALL(X) (X).begin(), (X).end()
#define REP(i,n) for (int i = 0; i < n; ++i)
#define DO(X) while(X--)
#define DRI(X) int (X); cin >> X
#define DRII(X, Y) int X, Y; cin >> X >> Y
#define DRIII(X, Y, Z) int X, Y, Z; cin >> X >> Y >> Z
#define read(X) for(int &ele : (X)) cin >> ele
#define readI(X) for(int idx = 1; idx < SZ((X)); ++idx) cin >> (X)[idx]
#define debug(x) cerr << '>' << #x << ':' << x << endl
#define MP make_pair
#define PB push_back
#define F first
#define S second

using namespace std;
typedef long long LL;
const int INF = 1<<29;
inline int in() { int x; cin >> x; return x; }
template<class T> void max_self(T & a, const T & b) { if(a < b) a = b; }
template<class T> void min_self(T & a, const T & b) { if(a > b) a = b; }

int main(){
	FAST_IO 
	LL taka, aoki, K; cin >> taka >> aoki >> K;
	LL takaS = min(taka, K);
	taka -= takaS;
	K -= takaS;
	aoki = max(0LL, aoki - K);
	cout << taka << " " << aoki << "\n";
	return 0;
}