Problem: Apart from having lots of holidays throughout the year, residents of Berland also have whole lucky years. Year is considered lucky if it has no more than 1 non-zero digit in its number. So years 100, 40000, 5 are lucky and 12, 3001 and 12345 are not.

You are given current year in Berland. Your task is to find how long will residents of Berland wait till the next lucky year.

Input

The first line contains integer number n (1 ≤ n ≤ 109) — current year in Berland.

Output

Output amount of years from the current year to the next lucky one.

Examples

Input

4


Output

1


Input

201


Output

99


Input

4000


Output

1000

Note

In the first example next lucky year is 5. In the second one — 300. In the third — 5000.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long inline gcd(long long x, long long y) {
  return !y ? (long long)abs(x) : gcd(y, x % y);
}
long long inline lcm(long long a, long long b) {
  return ((long long)abs((a / gcd(a, b)) * b));
}
double inline cartesian_dist(double x1, double y1, double x2, double y2) {
  return sqrt((x1 - x2) * (x1 - x2) + (y1 - y2) * (y1 - y2));
}
long long bigmod(long long b, long long p, long long m) {
  if (!p)
    return 1 % m;
  else if (p & 1)
    return (((b % m) + m) % m * bigmod(b, p - 1, m)) % m;
  else {
    long long ret = bigmod(b, p / 2, m);
    return (ret * ret) % m;
  }
}
long long inverse(long long a, long long m) { return bigmod(a, m - 2, m); }
long long extended(long long a, long long m) {
  long long oldr = a, r = m, oldans = 1, ans = 0, quotient;
  while (r) {
    quotient = oldr / r;
    oldr = oldr - r * quotient;
    swap(oldr, r);
    oldans = oldans - ans * quotient;
    swap(oldans, ans);
  }
  return ((oldans % m) + m) % m;
}
const double PI = acos(-1.0);
const double eps = 10E-12;
int main() {
  long long n;
  scanf("%lld", &n);
  long long temp = n, cnt = 0, mul = 1;
  while (temp > 9) {
    temp /= 10;
    mul *= 10;
  }
  cout << (temp + 1) * mul - n;
  return 0;
}