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;
template <class T>
inline T bigmod(T p, T e, T M) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class T>
inline T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <class T>
inline T lcm(T a, T b) {
  return (a / gcd(a, b)) * b;
}
template <class T>
inline T modinverse(T a, T M) {
  return bigmod(a, M - 2, M);
}
template <class T>
inline T bpow(T p, T e) {
  long long ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p);
    p = (p * p);
  }
  return (T)ret;
}
template <class T>
inline T gethash(T h[], T power[], int l, int r) {
  if (r < l) return (T)0;
  return h[l] - h[r + 1] * power[r - l + 1];
}
struct data {
  data() {}
  bool friend operator<(data a, data b) { return 0; }
};
int n, m, k;
long long N, M, K;
int main() {
  int t, tc = 1;
  scanf("%d %d", &n, &m);
  string s1 = "";
  string s2 = "";
  string s = "";
  s = "1";
  for (int i = 1; i <= 1200; i++) {
    cout << 1;
  }
  cout << '\n';
  for (int i = 1; i <= 1199; i++) {
    cout << 8;
  }
  cout << 9 << '\n';
  cout << s2 << '\n';
  return 0;
}