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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <cstdlib>
#include <time.h>
#include <algorithm>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <array>
#include <string>
#include <iomanip>
#include <cstring>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define mp make_pair
#define eb emplace_back
#define f first
#define s second
#define sz(x) (ll)x.size()
#define lnt(s) (ll)s.length()
#define mt(cnt) cnt.empty()
using namespace std;
using namespace __gnu_pbds;

typedef long long int ll;
typedef tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update> osSet;

using pll = pair<ll, ll>;
using pl = pair<ll, pll>;
using tll = tuple<ll, ll, ll>;

const ll maxN = 1e5;
const ll INF = 1e17;
const ll MOD = 1e9 + 7;

ll xs[] = {-1, 1, 0, 0, -1, 1, 1, -1};
ll ys[] = {0, 0, -1, 1, -1, 1, -1, 1};

ll segtree[maxN << 2];
ll lazy[maxN << 2];
ll A[maxN];

inline ll left(ll x){
	return x << 1;
}

inline ll right(ll x){
	return (x<<1) + 1;
}

ll func(ll x){
	return (x*(x+1))/2;
}

ll _pow(ll b, ll p){
	if(p == 0) return 1;
	if(p == 1) return b;
	ll exB = b*b;
	if(p&1) return b*_pow(exB, p/2);
	return _pow(exB, p/2);
}

int main(){
	cin.sync_with_stdio(false);
	cin.tie(0);
	ll a, b, k;
	cin >> a >> b >> k;
	cout << (a-k <= 0 ? 0 : a-k) << " " << (a-k < 0 ? b-abs(a-k) : b);
	return 0;
}