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;
void dfs(vector<int> adj[], bool visited[], int u, int dp[]) {
  visited[u] = 1;
  int mx = 0;
  for (int i = 0; i < adj[u].size(); i++)
    if (!visited[adj[u][i]]) {
      dfs(adj, visited, adj[u][i], dp);
      mx = max(mx, dp[adj[u][i]]);
    }
  dp[u] = mx + 1;
}
struct s {
  long long int l, r;
};
bool compare1(s s1, s s2) {
  if (s1.l > s2.l)
    return 1;
  else if (s1.l == s2.l)
    return (s1.r <= s2.r);
  else
    return 0;
}
long long int gcd(long long int a, long long int b) {
  if (a == 0)
    return b;
  else
    return gcd(b % a, a);
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= 2228; i++) cout << "2";
  cout << " ";
  for (int i = 1; i <= 2227; i++) cout << "7";
  cout << "8";
  return 0;
}
