Problem: Vasya has a non-negative integer n. He wants to round it to nearest integer, which ends up with 0. If n already ends up with 0, Vasya considers it already rounded.

For example, if n = 4722 answer is 4720. If n = 5 Vasya can round it to 0 or to 10. Both ways are correct.

For given n find out to which integer will Vasya round it.

Input

The first line contains single integer n (0 ≤ n ≤ 109) — number that Vasya has.

Output

Print result of rounding n. Pay attention that in some cases answer isn't unique. In that case print any correct answer.

Examples

Input

5


Output

0


Input

113


Output

110


Input

1000000000


Output

1000000000


Input

5432359


Output

5432360

Note

In the first example n = 5. Nearest integers, that ends up with zero are 0 and 10. Any of these answers is correct, so you can print 0 or 10.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long fast_expo(long long base, long long exp, long long module) {
  long long ans = 1;
  while (exp) {
    if (exp & 1) ans = (ans * base) % module;
    base = (base * base) % module;
    exp >>= 1;
  }
  return ans;
}
bool primes[100001];
void seive() {
  memset(primes, true, sizeof(primes));
  primes[0] = false;
  primes[1] = false;
  for (int i = 2; i * i <= 100000; i++) {
    if (primes[i]) {
      for (int j = i * i; j <= 100000; j += i) primes[j] = false;
    }
  }
}
bool isprime(int n) {
  if (n == 1) return false;
  if (n == 2) return true;
  if (n % 2 == 0) return false;
  for (int i = 3; i * i <= n; i += 2)
    if (n % i == 0) return false;
  return true;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long f[100001];
long long fib(long long n) {
  if (n == 0) return 0;
  if (n == 1 || n == 2) return (f[n] = 1);
  if (f[n]) return f[n];
  long long k = (n & 1) ? (n + 1) / 2 : n / 2;
  f[n] = (n & 1) ? (fib(k) * fib(k) + fib(k - 1) * fib(k - 1))
                 : (2 * fib(k - 1) + fib(k)) * fib(k);
  return f[n];
}
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  int t1 = n % 10;
  int t2 = t1;
  int cnt1 = 0, cnt2 = 0;
  if (t1 == 0) {
    cout << n << endl;
    return 0;
  }
  while (t1 <= 10) {
    cnt1++;
    t1++;
  }
  while (t2 > 0) {
    t2--;
    cnt2++;
  }
  if (cnt1 <= cnt2)
    cout << (n + cnt1 - 1) << endl;
  else
    cout << (n - cnt2) << endl;
  return 0;
}