#include <bits/stdc++.h>
using namespace std;
int dp[100];
struct point {
 public:
  long long x, y;
  point(long long a, long long b) {
    x = a;
    y = b;
  }
  point() {}
  void setPoint(long long a, long long b) {
    x = a;
    y = b;
  }
  bool operator<(const point &p) const {
    return x < p.x || (x == p.x && y < p.y);
  }
};
long long msum(long long x) {
  long long cdg = x % 10, pdg = cdg + 1;
  x -= cdg;
  long long s = dp[cdg], p = 10;
  while (x) {
    cdg = x % (p * 10);
    x -= cdg;
    cdg /= p;
    if (cdg) s += (dp[cdg - 1] + cdg * 45) * (p / 10) + dp[pdg] * cdg;
    p *= 10;
    pdg = cdg + 1;
  }
  return s;
}
long long sum(long long a) {
  if (a == 0) return 0;
  int digit = 0;
  long long b = a;
  while (b >= 10) {
    digit++;
    b /= 10;
  }
  int b_digit = b;
  long long ans = b * digit;
  for (int i = 0; i < digit; i++) ans *= 10, b *= 10;
  ans *= 9;
  ans /= 2;
  for (int i = 0; i < b_digit; i++) ans += i * (b / b_digit);
  ans += sum(a - b) + b_digit * (a - b + 1);
  return ans;
}
long long dsum(long long x) {
  long long s = 0, p = 10;
  while (x) {
    long long d = x % p;
    x -= d;
    d *= 10;
    d /= p;
    p *= 10;
    s += d;
  }
  return s;
}
int main() {
  long long ans = 0;
  long long X, Y;
  cin >> X;
  memset(dp, 0, sizeof(dp));
  for (int i = (1); i < (10); i++) dp[i] = dp[i - 1] + i;
  long long lo = 0, mi, hi = 20000000000000000;
  while (lo < hi) {
    mi = (lo + hi) >> 1;
    Y = sum(mi);
    if (Y < X)
      lo = mi + 1;
    else
      hi = mi;
  }
  lo = 1;
  long long s = sum(hi) - sum(lo - 1);
  while (s != X) {
    if (s < X)
      s += dsum(++hi);
    else
      s -= dsum(lo++);
  }
  cout << lo << " " << hi << endl;
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

Little X has met the following problem recently. 

Let's define f(x) as the sum of digits in decimal representation of number x (for example, f(1234) = 1 + 2 + 3 + 4). You are to calculate <image>

Of course Little X has solved this problem quickly, has locked it, and then has tried to hack others. He has seen the following C++ code: 
    
    
      
        ans = solve(l, r) % a;  
        if (ans <= 0)  
          ans += a;  
      
    

This code will fail only on the test with <image>. You are given number a, help Little X to find a proper test for hack.

Input

The first line contains a single integer a (1 ≤ a ≤ 1018).

Output

Print two integers: l, r (1 ≤ l ≤ r < 10200) — the required test data. Leading zeros aren't allowed. It's guaranteed that the solution exists.

Examples

Input

46


Output

1 10


Input

126444381000032


Output

2333333 2333333333333

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int dp[100];
struct point {
 public:
  long long x, y;
  point(long long a, long long b) {
    x = a;
    y = b;
  }
  point() {}
  void setPoint(long long a, long long b) {
    x = a;
    y = b;
  }
  bool operator<(const point &p) const {
    return x < p.x || (x == p.x && y < p.y);
  }
};
long long msum(long long x) {
  long long cdg = x % 10, pdg = cdg + 1;
  x -= cdg;
  long long s = dp[cdg], p = 10;
  while (x) {
    cdg = x % (p * 10);
    x -= cdg;
    cdg /= p;
    if (cdg) s += (dp[cdg - 1] + cdg * 45) * (p / 10) + dp[pdg] * cdg;
    p *= 10;
    pdg = cdg + 1;
  }
  return s;
}
long long sum(long long a) {
  if (a == 0) return 0;
  int digit = 0;
  long long b = a;
  while (b >= 10) {
    digit++;
    b /= 10;
  }
  int b_digit = b;
  long long ans = b * digit;
  for (int i = 0; i < digit; i++) ans *= 10, b *= 10;
  ans *= 9;
  ans /= 2;
  for (int i = 0; i < b_digit; i++) ans += i * (b / b_digit);
  ans += sum(a - b) + b_digit * (a - b + 1);
  return ans;
}
long long dsum(long long x) {
  long long s = 0, p = 10;
  while (x) {
    long long d = x % p;
    x -= d;
    d *= 10;
    d /= p;
    p *= 10;
    s += d;
  }
  return s;
}
int main() {
  long long ans = 0;
  long long X, Y;
  cin >> X;
  memset(dp, 0, sizeof(dp));
  for (int i = (1); i < (10); i++) dp[i] = dp[i - 1] + i;
  long long lo = 0, mi, hi = 20000000000000000;
  while (lo < hi) {
    mi = (lo + hi) >> 1;
    Y = sum(mi);
    if (Y < X)
      lo = mi + 1;
    else
      hi = mi;
  }
  lo = 1;
  long long s = sum(hi) - sum(lo - 1);
  while (s != X) {
    if (s < X)
      s += dsum(++hi);
    else
      s -= dsum(lo++);
  }
  cout << lo << " " << hi << endl;
  return 0;
}

```