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;
long long gcd(long long a, long long b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long Max(long long a, long long b) {
  if (a > b) return a;
  return b;
}
long long Min(long long a, long long b) {
  if (a < b) return a;
  return b;
}
long long gcdExtended(long long a, long long b, long long *x, long long *y) {
  if (a == 0) {
    *x = 0, *y = 1;
    return b;
  }
  long long x1, y1;
  long long gcd = gcdExtended(b % a, a, &x1, &y1);
  *x = y1 - (b / a) * x1;
  *y = x1;
  return gcd;
}
long long modInverse(long long a, long long m) {
  long long x, y;
  long long g = gcdExtended(a, m, &x, &y);
  if (g != 1)
    return -1;
  else {
    long long res = (x % m + m) % m;
    return res;
  }
}
vector<int> primeFactors(int n) {
  vector<int> v;
  while (n % 2 == 0) {
    v.push_back(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      v.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) v.push_back(n);
  return v;
}
long long binarySearch(long long arr[], long long l, long long r, long long x) {
  if (r >= l) {
    long long mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
long long func(long long num) {
  long long res = 0;
  while (num) {
    res += num % 10;
    num = num / 10;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, m;
  cin >> n >> m;
  long long ninc = n / 9;
  long long rem = n % 9;
  long long a;
  long long b;
  long long c = 0;
  long long a1;
  long long a2;
  long long falg = 0;
  for (long long i = 1; i <= 2229; i++) {
    cout << 5;
  }
  cout << endl;
  for (long long i = 1; i <= 2229; i++) {
    if (i == 2229)
      cout << 5;
    else
      cout << 4;
  }
  return 0;
}