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 dx44[5] = {0, -1, -1, 1, 1};
const int dy44[5] = {0, -1, 1, -1, 1};
const int dx4[5] = {0, 1, 0, -1};
const int dy4[5] = {1, 0, -1, 0};
const int dx8[9] = {0, -1, 0, 1, 0, 1, 1, -1, -1};
const int dy8[9] = {0, 0, 1, 0, -1, 1, -1, 1, -1};
const int dx82[9] = {0, -1, -1, 1, 1, 2, 2, -2, -2};
const int dy82[9] = {0, 2, -2, 2, -2, 1, -1, 1, -1};
const int maxn = 5e5 + 50;
const double PI = acos(-1.0);
const long long mod = 1e9 + 7;
long long po(long long a, long long b, long long mod) {
  long long res = 1;
  a %= mod;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  } else {
    return gcd(b % a, a);
  }
}
void extgcd(long long a, long long b, long long& d, long long& x,
            long long& y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    extgcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inverse(long long a, long long n) {
  long long d, x, y;
  extgcd(a, n, d, x, y);
  return d == 1 ? (x + n) % n : -1;
}
void YES() {
  puts("YES");
  exit(0);
}
void Yes() { puts("Yes"); }
void NO() {
  puts("NO");
  exit(0);
}
void No() { puts("No"); }
void one() {
  puts("-1");
  exit(0);
}
void solve() {
  for (int i = 1; i <= 2228; i++) {
    cout << "4";
  }
  cout << "6" << endl;
  for (int i = 1; i <= 2228; i++) {
    cout << "5";
  }
  cout << "4" << endl;
}
int main() {
  int t;
  t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
