Let s(x) be sum of digits in decimal representation of positive integer x. Given two integers n and m, find some positive integers a and b such that 

  * s(a) ≥ n, 
  * s(b) ≥ n, 
  * s(a + b) ≤ m. 

Input

The only line of input contain two integers n and m (1 ≤ n, m ≤ 1129).

Output

Print two lines, one for decimal representation of a and one for decimal representation of b. Both numbers must not contain leading zeros and must have length no more than 2230.

Examples

Input

6 5


Output

6 
7


Input

8 16


Output

35 
53

Note

In the first sample, we have n = 6 and m = 5. One valid solution is a = 6, b = 7. Indeed, we have s(a) = 6 ≥ n and s(b) = 7 ≥ n, and also s(a + b) = s(13) = 4 ≤ m.
#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9 + 7;
inline void rvi(vector<int>& v, int n) {
  int temp;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    v.push_back(temp);
  }
}
inline void rvll(vector<long long int>& v, int n) {
  long long int temp;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    v.push_back(temp);
  }
}
inline void prvi(vector<int>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
inline void prvll(vector<long long int>& a) {
  for (int i = 0; i < a.size(); i++) {
    cout << a[i] << " ";
  }
  cout << endl;
}
inline void prmii(map<int, int>& m) {
  for (auto x : m) {
    cout << x.first << " " << x.second << endl;
  }
}
inline vector<long long int> prefixsum(vector<int>& a) {
  vector<long long int> b;
  b.push_back(a[0]);
  for (int i = 1; i < a.size(); i++) {
    b.push_back(a[i] + b[i - 1]);
  }
  return b;
}
inline vector<long long int> suffixsum(vector<int>& a) {
  vector<long long int> b((a.size()));
  b[(a.size()) - 1] = a[(a.size()) - 1];
  for (int i = (a.size()) - 1 - 1; i >= 0; i--) {
    b[i] = a[i] + b[i + 1];
  }
  return b;
}
inline vector<long long int> prefixsum(vector<long long int>& a) {
  vector<long long int> b;
  b.push_back(a[0]);
  for (int i = 1; i < a.size(); i++) {
    b.push_back(a[i] + b[i - 1]);
  }
  return b;
}
inline vector<long long int> suffixsum(vector<long long int>& a) {
  vector<long long int> b((a.size()));
  b[(a.size()) - 1] = a[(a.size()) - 1];
  for (int i = (a.size()) - 1 - 1; i >= 0; i--) {
    b[i] = a[i] + b[i + 1];
  }
  return b;
}
void solve();
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  return 0;
}
void solve() {
  int n, m;
  cin >> n >> m;
  string s1, s2;
  for (int i = 0; i < 150; i++) s1.insert(i, 1, '9');
  for (int i = 0; i < 150; i++) s1.insert(i + 150, 1, '0');
  s1.insert(300, 1, '1');
  for (int i = 0; i < 150; i++) s2.insert(i, 1, '9');
  s2.insert(150, 1, '9');
  cout << s1 << endl << s2 << endl;
}
