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: Wrong Answer)

#include<iostream>
#include<bits/stdc++.h> 
#include<string>
#define UB upper_bound
#define LB lower_bound
#define BS binary_search
#define IN insert
#define PB push_back
#define EB emplace_back
#define MP make_pair
#define NL cout<<endl
#define ll long long int
#define ld long double
#define vl vector<ll>
#define sl set<ll>
#define ml map<ll,ll>
#define sc set<char>
#define li list<ll>
#define vp vector<pair<ll, ll> >
#define vs vector<string>
#define ss set<string>
#define REP(i,a,b) for(ll i=a;i<b;i++)
#define REPI(i,a,b) for(ll i=b-1;i>=a;i--)
#define REPIT(it,a) for(auto it=a.begin();it!=a.end();it++)
#define N (ll)pow(2,32)-1
#define mod (ll)1000000007
using namespace std;
vl f;
void sieve(ll m){
  REP(i,1,sqrt(m)+1){
    if(m%i==0){
      f.PB(i);
      if(m/i!=i)
       f.PB(m/i);
    }
  }
}
int main()
{

ios::sync_with_stdio(0);
cin.tie(0);
ll t=1;
//cin>>t;
while(t--){
 ll a,b,k; cin>>a>>b>>k;
 cout<<max(a-k,0ll)<<" "<<max(a+b-k,0ll)<<endl;
}
return 0; 
}