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;
int ss(int x) {
  int ans = 0;
  while (x) {
    ans += x % 10;
    x /= 10;
  }
  return ans;
}
int main() {
  int n, m;
  cin >> n >> m;
  int quo = m / 9;
  int rem = m % 9;
  vector<int> vc;
  int cnt = 0;
  for (int i = 0; i < quo; i++) {
    vc.push_back(9);
    cnt++;
  }
  if (rem) {
    vc.push_back(rem);
    cnt++;
  }
  cnt = 2000 - cnt;
  for (int i = 0; i < cnt; i++) {
    vc.push_back(0);
  }
  cnt = 2000 - n;
  vector<int> gc;
  for (int i = 0; i < cnt; i++) {
    gc.push_back(0);
  }
  for (int i = 0; i < n; i++) {
    gc.push_back(1);
  }
  for (int i = 1999; i > 0; i--) {
    if (vc[i] >= gc[i]) {
      vc[i] = vc[i] - gc[i];
    } else {
      vc[i - 1]--;
      vc[i] = (10 + vc[i] - gc[i]);
    }
  }
  for (int i = 0; i < n; i++) {
    cout << 1;
  }
  cout << " ";
  int i;
  for (i = 0; i < 2000; i++) {
    if (vc[i] != 0) break;
  }
  for (int j = i; j < 2000; j++) {
    cout << vc[j];
  }
}
