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 MAX = 2e5 + 10;
long long tab[MAX];
long long s(string &a) {
  long long result = 0;
  for (int i = 0; i < a.size(); i++) {
    result += a[i] - '0';
  }
  return result;
}
int main() {
  ios_base::sync_with_stdio(0);
  long long n, m;
  cin >> n >> m;
  if (n == 1 && m == 1) {
    cout << "1\n9";
    return 0;
  }
  string a = "", b = "";
  for (int i = 0; i < n; i++) a += '1';
  long long diff = m - n;
  string prefix = "", sufix = "";
  long long value = 0;
  if (diff > 0) {
    for (int i = 0; i < diff; i++) {
      b += '1';
      value++;
    }
    for (int i = 0; i < n; i++) b += '0';
    while (value < n) {
      prefix += '1';
      sufix += '8';
      value += 9;
    }
    sufix += '9';
  } else if (diff == 0) {
    b += "8";
    value += 17;
    for (int i = 0; i < n - 2; i++) b += '0';
    b += '1';
    value++;
    while (value < n) {
      prefix += '1';
      sufix += '8';
      value += 9;
    }
    sufix += '9';
    reverse(b.begin(), b.end());
  } else {
    for (int i = 0; i < -diff + 1; i++) {
      b += '8';
      value += 8;
    }
    for (int i = 0; i < n + diff - 3; i++) b += '0';
    b += '1';
    value++;
    if (-diff == n - 1) {
      b[b.size() - 2] = '9';
      b[b.size() - 1] = '9';
      value += 9;
    }
    while (value < n) {
      prefix += '1';
      sufix += '8';
      value += 9;
    }
    sufix += '9';
    reverse(b.begin(), b.end());
  }
  cout << a << endl << prefix << b << sufix << endl;
}
