codecodefor 808A——Lucky Year

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本文介绍了一个算法问题，旨在找出从当前年份到下一个幸运年份所需的年数。幸运年份定义为该年份数字中最多只有一个非零数字的年份。文章提供了示例输入输出及一个简单的算法实现。

A. Lucky Year

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <map>
#include <set>
using namespace std;

#define _ sync_with_stdio(false)
const int MAXN=100000+10;
const int INF=0x7fffffff;

long long n;
//bool c(int n){
//	int num=0;
//	while(n){
//		if(n%10!=0){
//			num++;
//		}
//		n/=10;
//		if(num>1)
//			return false;
//	}
//	return true;
//}
int main() {
	cin>>n;
	long long temp=n;
	long long t=1;
	long long x;
	while(temp){
		x=temp%10;
		temp/=10;
		t*=10;
	}
	t/=10;
	cout<<(x+1)*t-n<<endl;
}