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;
typedef pair<int, int> PII;
long long qpow(long long a, long long b, long long m) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % m;
    a = a * a % m, b >>= 1;
  }
  return ans;
}
long long qpow(long long a, long long b) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a, b >>= 1;
  }
  return ans;
}
long long qmul(long long a, long long b, long long m) {
  long long res = 0;
  while (b > 0) {
    if (b & 1) res = (res + a) % m;
    a = (a + a) % m;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long inv(long long t, long long p) {
  return t == 1 ? 1 : (p - p / t) * inv(p % t, p) % p;
}
long long exgcd(long long a, long long b, long long &x, long long &y) {
  if (b == 0) {
    x = 1;
    y = 0;
    return a;
  }
  long long r = exgcd(b, a % b, y, x);
  y -= (a / b) * x;
  return r;
}
long long inv1(long long a, long long p) {
  long long d, x, y;
  d = exgcd(a, p, x, y);
  return d == 1 ? (x % p + p) % p : -1;
}
template <typename T, typename B>
T Add(T a, B b, long long m) {
  return a + b >= m ? (a + b) % m : a + b;
}
template <typename T, typename B>
T Sub(T a, B b, long long m) {
  return a - b < 0 ? (a - b + m) % m : a - b;
}
template <class T>
void _sf(T &x) {
  cin >> x;
}
void _sf(int &x) { scanf("%d", &x); }
void _sf(long long &x) { scanf("%lld", &x); }
void _sf(double &x) { scanf("%lf", &x); }
void _sf(char &x) { scanf(" %c", &x); }
void _sf(char *x) { scanf("%s", x); }
void sf() {}
template <class T, class... U>
void sf(T &head, U &...tail) {
  _sf(head);
  sf(tail...);
}
template <class T>
void _pf(const T &x) {
  cout << x;
}
void _pf(const int &x) { printf("%d", x); }
void _pf(const long long &x) { printf("%lld", x); }
void _pf(const double &x) { printf("%.16f", x); }
void _pf(const char &x) { putchar(x); }
void _pf(const char *x) { printf("%s", x); }
template <class T, class U>
void _pf(const pair<T, U> &x) {
  pf(x.first);
  putchar(' ');
  pf(x.second);
}
template <class T>
void _pf(const vector<T> &x) {
  for (auto i = x.begin(); i != x.end(); _pf(*i++))
    if (i != x.cbegin()) putchar(' ');
}
void pf() {}
template <class T, class... U>
void pf(const T &head, const U &...tail) {
  _pf(head);
  putchar(sizeof...(tail) ? ' ' : '\n');
  pf(tail...);
}
template <typename T>
inline void read(T &x) {
  x = 0;
  static int p;
  p = 1;
  static char c;
  c = getchar();
  while (!isdigit(c)) {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 1) + (x << 3) + (c - 48);
    c = getchar();
  }
  x *= p;
}
const bool IOS = false;
const double PI = acos(-1);
const double eps = 1e-8;
const long long linf = 0x3f3f3f3f3f3f3f3f;
const long long INV2 = 500000004;
const long long mod = 1e9 + 7;
const int inf = 0x3f3f3f3f;
const int maxn = 200000 + 10;
char s[maxn];
int ans[maxn];
vector<char> mx;
void Main();
int main() {
  if (IOS) std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  Main();
  0;
  return 0;
}
void Main() {
  int n;
  sf(n);
  if (n % 10 <= 5)
    pf(n - n % 10);
  else
    pf(n - n % 10 + 10);
}