Problem: 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.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int N, M;
void solve() {
  vector<int> A(1, 5), B(1, 5);
  int sumA = 5, sumB = 5;
  int i = 0;
  int D[2] = {4, 5};
  while (sumA < N || sumB < N) {
    A.push_back(D[i % 2]);
    B.push_back(D[(i % 2) ^ 1]);
    ++i;
    sumA += A.back(), sumB += B.back();
  }
  reverse(A.begin(), A.end());
  reverse(B.begin(), B.end());
  string SA = "", SB = "";
  for (int i = 0; i < (int((A).size())); ++i) SA += '0' + A[i];
  for (int i = 0; i < (int((B).size())); ++i) SB += '0' + B[i];
  assert(int((SA).size()) <= 2230 && int((SB).size()) <= 2230);
  cout << SA << " " << SB << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  solve();
  return 0;
}