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 JAVA):

Solution:
//var print = require('./print');
//var readline = require('./readline');

var n = parseInt(readline());
print(Math.round(n/10)*10);
/*
var args = readline().split(' ');

var n = parseInt(args[0]);
var m = parseInt(args[1]);
var k = parseInt(args[2]);

args = readline().split(' ');
var a = [];
for(var i=0;i<n;i++){
  a.push(parseInt(args[i]));
};
a = a.sort(function(a,b){
  return a-b;
})
var pp = a[a.length-1] + 1;
var b = {};
for(var i=1;i<pp;i++){
  b[i] = [];
}
for(var i=0;i<n;i++){
  var current = a[i];
  for(var j = current;j > 0 && j>current - m;j--){
    b[j].push(current);
  }
};
print(b);
print('----------');
var count = 0;
for(var i=1;i<pp;i++){
  if(b[i].length>=k){
    var current = b[i][b[i].length - 1];
    for(var j = i; j < i + Math.max(m,current); j++){
      var index = b[j].indexOf(current);
      if(index > -1)
        b[j].splice(index,1);
    }
    print(current);
    count ++;
    if(b[i].length >= k){
      i--;
    }
  }
}
print(b);
print('--------');
print(count);
*/
/* A
var n = readline();
var res = '';
for (var i = 0; i < n.length - 2; i++)
  res += n[i];
if(n.length > 1){
  if(n[n.length - 1] > '5')
    res += parseInt(n[n.length - 2]) + 1;
  else {
    res += n[n.length - 2];
  }
}
res += '0';
print(res);
*/