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;
void fastio() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
}
template <typename T>
T __lcm(T a, T b) {
  return (a * b) / __gcd(a, b);
}
template <typename T>
T fpow(T a, T b, T p) {
  a %= p;
  T res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return res;
}
template <typename T>
T binpow(T a, T b) {
  T res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
const long double PI = acos(-1);
const long long MOD = 1000000000 + 7;
const int maxn = 2230;
int n, m;
vector<int> a, b, s;
int main() {
  fastio();
  cin >> n >> m;
  int curs = 5;
  b.push_back(5);
  while (curs < n) b.push_back(4), curs += 4;
  reverse((b).begin(), (b).end());
  a.resize(b.size(), 5);
  for (auto& v : a) printf("%d", v);
  puts("");
  for (auto& v : b) printf("%d", v);
  return 0;
}
