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>
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const long double PI = acos(-1);
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
const int mod1 = 998244353;
const long long INF = 2e18;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
int read() {
  long long s = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    s = s * 10 + c - '0';
    c = getchar();
  }
  return s * f;
}
long long Pow(long long a, long long b) {
  if (a == 1 || b == 1) return a;
  if (b % 2 == 1) return (a * Pow(a, b - 1)) % MOD;
  return (Pow(a, b / 2) * Pow(a, b / 2)) % MOD;
}
int gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool isPrime(int n) {
  if (n <= 1) return false;
  for (int i = 2; i < n; i++)
    if (n % i == 0) return false;
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n - 1; i++) {
      cout << 8;
    }
    cout << 9;
    cout << "\n";
    for (int i = 0; i < n; i++) {
      cout << 1;
    }
  }
  return 0;
}
