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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<map>
#include<set>
#include<cstdio>
#include<cmath>
#include<deque>
#include<numeric>
#include<queue>
#include<stack>
#include<cstring>
#include<limits>
#include<functional>
#include<unordered_set>
#include<iomanip>
#include<cassert>
#include<regex>
#include<bitset>
#include<complex>
#include<chrono>
#define rep(i,a) for(int i=(int)0;i<(int)a;++i)
#define pb push_back
#define eb emplace_back
#define all(x) x.begin(),x.end()
using ll=long long;
constexpr ll mod = 1e9 + 7;
constexpr ll INF = 1LL << 60;

ll gcd(ll n, ll m) {
	ll tmp;
	while (m!=0) {
		tmp = n % m;
		n = m;
		m = tmp;
	}
	return n;
}

ll lcm(ll n, ll m) {
	return abs(n * m) / gcd(n, m);//gl=xy
}
 
template<class T> inline bool chmin(T& a, T b) {
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template<class T> inline bool chmax(T& a, T b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
 
using namespace std;

//ここから
template< class T >
struct Matrix {
  vector< vector< T > > A;

  Matrix() {}

  Matrix(size_t n, size_t m) : A(n, vector< T >(m, 0)) {}

  Matrix(size_t n) : A(n, vector< T >(n, 0)) {};

  size_t height() const {
    return (A.size());
  }

  size_t width() const {
    return (A[0].size());
  }

  inline const vector< T > &operator[](int k) const {
    return (A.at(k));
  }

  inline vector< T > &operator[](int k) {
    return (A.at(k));
  }

  static Matrix I(size_t n) {
    Matrix mat(n);
    for(int i = 0; i < n; i++) mat[i][i] = 1;
    return (mat);
  }

  Matrix &operator+=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() && m == B.width());
    for(int i = 0; i < n; i++)
      for(int j = 0; j < m; j++)
        (*this)[i][j] += B[i][j];
    return (*this);
  }

  Matrix &operator-=(const Matrix &B) {
    size_t n = height(), m = width();
    assert(n == B.height() && m == B.width());
    for(int i = 0; i < n; i++)
      for(int j = 0; j < m; j++)
        (*this)[i][j] -= B[i][j];
    return (*this);
  }

  Matrix &operator*=(const Matrix &B) {
    size_t n = height(), m = B.width(), p = width();
    assert(p == B.height());
    vector< vector< T > > C(n, vector< T >(m, 0));
    for(int i = 0; i < n; i++)
      for(int j = 0; j < m; j++)
        for(int k = 0; k < p; k++)
          C[i][j] = (C[i][j] + (*this)[i][k] * B[k][j]);
    A.swap(C);
    return (*this);
  }

  Matrix &operator^=(long long k) {
    Matrix B = Matrix::I(height());
    while(k > 0) {
      if(k & 1) B *= *this;
      *this *= *this;
      k >>= 1LL;
    }
    A.swap(B.A);
    return (*this);
  }

  Matrix operator+(const Matrix &B) const {
    return (Matrix(*this) += B);
  }

  Matrix operator-(const Matrix &B) const {
    return (Matrix(*this) -= B);
  }

  Matrix operator*(const Matrix &B) const {
    return (Matrix(*this) *= B);
  }

  Matrix operator^(const long long k) const {
    return (Matrix(*this) ^= k);
  }

  friend ostream &operator<<(ostream &os, Matrix &p) {
    size_t n = p.height(), m = p.width();
    for(int i = 0; i < n; i++) {
      os << "[";
      for(int j = 0; j < m; j++) {
        os << p[i][j] << (j + 1 == m ? "]\n" : ",");
      }
    }
    return (os);
  }


  T determinant() {
    Matrix B(*this);
    assert(width() == height());
    T ret = 1;
    for(int i = 0; i < width(); i++) {
      int idx = -1;
      for(int j = i; j < width(); j++) {
        if(B[j][i] != 0) idx = j;
      }
      if(idx == -1) return (0);
      if(i != idx) {
        ret *= -1;
        swap(B[i], B[idx]);
      }
      ret *= B[i][i];
      T vv = B[i][i];
      for(int j = 0; j < width(); j++) {
        B[i][j] /= vv;
      }
      for(int j = i + 1; j < width(); j++) {
        T a = B[j][i];
        for(int k = 0; k < width(); k++) {
          B[j][k] -= B[i][k] * a;
        }
      }
    }
    return (ret);
  }
};

template< int mod >
struct ModInt {
  int x;

  ModInt() : x(0) {}

  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

  ModInt &operator+=(const ModInt &p) {
    if((x += p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator-=(const ModInt &p) {
    if((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }

  ModInt &operator*=(const ModInt &p) {
    x = (int) (1LL * x * p.x % mod);
    return *this;
  }

  ModInt &operator/=(const ModInt &p) {
    *this *= p.inverse();
    return *this;
  }

  ModInt operator-() const { return ModInt(-x); }

  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }

  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }

  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }

  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }

  bool operator==(const ModInt &p) const { return x == p.x; }

  bool operator!=(const ModInt &p) const { return x != p.x; }

  ModInt inverse() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while(b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }

  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while(n > 0) {
      if(n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }

  friend ostream &operator<<(ostream &os, const ModInt &p) {
    return os << p.x;
  }

  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt< mod >(t);
    return (is);
  }

  static int get_mod() { return mod; }
};

using modint = ModInt< mod >;



void solve(){
  ll a,b,k;
  cin>>a>>b>>k;
  ll aa=a;
  a=max(0LL,a-k);
  k-=aa;
  if(k>0)b=max(0LL,b-k);
  cout<<a<<" "<<b;
}

int main(){
	ios::sync_with_stdio(false);
  cin.tie(0);
	cout<<fixed<<setprecision(15);
  solve();
	return 0;
}