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 SORT(v, n) sort(v, v+n);
#define VSORT(v) sort(v.begin(), v.end());
#define INF 999999999
#define size_t unsigned long long
#define ll long long
#define rep(i,a) for(int i=0;i<(a);i++)
#define repr(i,a) for(int i=(int)(a)-1;i>=0;i--)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define FORR(i,a,b) for(int i=(int)(b)-1;i>=a;i--)
#define ALL(a) a.begin(), a.end()
using namespace std;
int si() { int x; scanf("%d", &x); return x; }
long long sl() { long long x; scanf("%lld", &x); return x; }
string ss() { string x; cin >> x; return x; }
void pi(int x) { printf("%d ", x); }
void pl(long long x) { printf("%lld ", x); }
void pd(double x) { printf("%.15f ", x); }
void ps(const string &s) { printf("%s ", s.c_str()); }
void br() { putchar('\n'); }

const int MOD = 1e9 + 7;

struct mint {
    int n;
    mint(int n_ = 0) : n(n_) {}
};

mint operator+(mint a, mint b) { a.n += b.n; if (a.n >= MOD) a.n -= MOD; return a; }
mint operator-(mint a, mint b) { a.n -= b.n; if (a.n < 0) a.n += MOD; return a; }
mint operator*(mint a, mint b) { return (long long)a.n * b.n % MOD; }
mint &operator+=(mint &a, mint b) { return a = a + b; }
mint &operator-=(mint &a, mint b) { return a = a - b; }
mint &operator*=(mint &a, mint b) { return a = a * b; }

typedef pair<int, int> P;

const int N = 1e6+5;
string s,t;
vector<ll> v;
ll a,b,k;

int main () {
  ios::sync_with_stdio(false); cin.tie(nullptr);
  cin >> a >> b >> k;
  if(a+b <= k) {
    cout << "0 0" << endl;
    return 0;
  }
  ll taka = a - k > 0 ? a-k : 0;
  ll aoki = a - k > 0 ? b : b+a-k;
  cout << taka << " " << aoki << endl;
}