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;
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = (ans * a);
    b = b >> 1;
    a = (a * a);
  }
  return ans;
}
long long int modInverse(long long int n, long long int m,
                         long long int mod = 1000000007) {
  return power(n, m - 2) % mod;
}
long long int sieve[1000001];
void factors(long long int num, set<long long int> &s) {
  if (num <= 1) return;
  s.insert(sieve[num]);
  factors(num / sieve[num], s);
}
long long int fun(char c) { return c - 'A' + 1; }
long long int visited[60][60];
long long int abcpath(char a[60][60], long long int n, long long int m,
                      long long int i, long long int j, long long int prev) {
  long long int x = 0;
  if (visited[i][j]) return 0;
  visited[i][j] = 1;
  if (i < n - 1 && prev == fun(a[i + 1][j]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i + 1, j, fun(a[i + 1][j])));
  if (j < m - 1 && prev == fun(a[i][j + 1]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i, j + 1, fun(a[i][j + 1])));
  if (i > 0 && prev == fun(a[i - 1][j]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i - 1, j, fun(a[i - 1][j])));
  if (j > 0 && prev == fun(a[i][j - 1]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i, j - 1, fun(a[i][j - 1])));
  if (i < n - 1 && j < m - 1 && prev == fun(a[i + 1][j + 1]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i + 1, j + 1, fun(a[i + 1][j + 1])));
  if (j < m - 1 && i > 0 && prev == fun(a[i - 1][j + 1]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i - 1, j + 1, fun(a[i - 1][j + 1])));
  if (i > 0 && j > 0 && prev == fun(a[i - 1][j - 1]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i - 1, j - 1, fun(a[i - 1][j - 1])));
  if (j > 0 && i < n - 1 && prev == fun(a[i + 1][j - 1]) - 1)
    x = max(x, 1 + abcpath(a, n, m, i + 1, j - 1, fun(a[i + 1][j - 1])));
  return x;
}
long long int sum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += n % 10;
    n /= 10;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int n, m;
  cin >> n >> m;
  long long int a1 = 0, a2 = 0;
  string s1 = "4", s2 = "5";
  while (a1 <= n) {
    s1 += '4';
    s2 += '5';
    a1 += 4;
  }
  s1 += '5';
  s2 += '5';
  cout << s1 << " " << s2 << endl;
  return 0;
}
