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 dir[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
const double PI = acos(-1.0), epx = 1e-6;
const long long mod = 1000003;
const int INF = 0x3f3f3f3f;
const int maxn = 1e6 + 10, N = 3010;
string s1, s2;
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    s1 = s2 = "";
    if (5 >= n) {
      printf("5\n5\n");
    } else {
      int num1 = 10, num2 = 9;
      s1 = "55";
      s2 = "54";
      while (num1 < n || num2 < n) {
        s1 += "5";
        s2 += "4";
        num1 += 5;
        num2 += 4;
      }
      cout << s1 << endl;
      for (int i = s2.size() - 1; i >= 0; --i) printf("%c", s2[i]);
      printf("\n");
    }
  }
  return 0;
}
