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 endl '\n'
#define fast ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define zy -2147382637
#define bql 2147483647
#define ll long long
#define ull unsigned long long
#define ld long double
#define il inline
#define sz(x) x.size()
#define maxn 100010
#define rp(i, l, r) for (int i = l; i <= r; i++)
#define rb(i, r, l) for (int i = r; i >= l; i--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair

using namespace std;
ll max(ll a,ll b){if(a>b)return a;else return b;}
ll min(ll a,ll b){if(a<b)return a;else return b;}
ll lowbit(ll x){return x&(-x);}
ll prime(ll x){
   if(x<=1)return false;
   for(int i=2;i<=int(sqrt(x));i++){
	   if(x%i==0)return false;
   }return true;
}
bool cmp(ll a,ll b){return a>b;}
ll gcd(ll a,ll b){ll r;while(b>0){r=a%b;a=b;b=r;}return a;}
ll powmod(ll a,ll b,ll mod) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
inline ll read() {
	char c = getchar(); ll w = 1; ll x = 0;
	while (!isdigit(c))
		(c == '-') && (w = -w), c = getchar();
	while (isdigit(c))
		x = (x << 1) + (x << 3) + (c ^ '0'), c = getchar();
	x *= w;
}
void fl(string name){
	freopen((name+".in").c_str(),"r",stdin);
	freopen((name+".out").c_str(),"w",stdout);
}
const int dx[8] = {-1, 1, 0, 0, -1, 1, -1, 1},
		  dy[8] = {0, 0, -1, 1, -1, -1, 1, 1};
long long n, m, k;
 
signed main()
{
	//freopen(".in", "r", stdin);
	//freopen(".out", "w", stdout);
	cin>>n>>m>>k;
	if (n <= k) 
	{
		k = k - n, printf("0 ");
		printf("%lld\n", max(1ll * 0, m - k));
	}
	else 
	{
		printf("%lld %lld\n", n - k, m);
	}
	return 0;
}