Problem:
Score : 200 points

Problem Statement

Kurohashi has never participated in AtCoder Beginner Contest (ABC).

The next ABC to be held is ABC N (the N-th ABC ever held).
Kurohashi wants to make his debut in some ABC x such that all the digits of x in base ten are the same.

What is the earliest ABC where Kurohashi can make his debut?

Constraints

100 \leq N \leq 999

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If the earliest ABC where Kurohashi can make his debut is ABC n, print n.

Sample Input 1

111

Sample Output 1

111

The next ABC to be held is ABC 111, where Kurohashi can make his debut.

Sample Input 2

112

Sample Output 2

222

The next ABC to be held is ABC 112, which means Kurohashi can no longer participate in ABC 111.
Among the ABCs where Kurohashi can make his debut, the earliest one is ABC 222.

Sample Input 3

750

Sample Output 3

777

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

//==================== default starts ====================//
#include <bits/stdc++.h>
 
using namespace std;
 
#define fread(x)        freopen(x,"r",stdin)
#define fwrite(x)       freopen (x,"w",stdout)
 
#define min3(a,b,c)     min(a,min(b,c))
#define max3(a,b,c)     max(a,max(b,c))
#define min4(a,b,c,d)   min(a,min(b,min(c,d)))
#define max4(a,b,c,d)   max(a,max(b,max(c,d)))
#define SQR(a)          ((a)*(a))
#define MEMSET(a,x)     memset(a,x,sizeof(a))
#define ABS(x)          ((x)<0?-(x):(x))
 
#define white           1
#define grey            2
#define black           3
 
#define ll              long long
#define pb              push_back
#define pf              push_front
#define INF             9e18
#define PI              acos(-1.0)
#define pii             pair<int,int>
#define pll             pair<ll, ll>
 
#define all(v)          v.begin(), v.end()
 
#define fastread        ios_base::sync_with_stdio(0);cin.tie(nullptr);
int on(int N,int pos){return N=N | (1<<pos);}
int off(int N,int pos){return N= N & ~(1<<pos);}
bool check(int N,int pos){return (bool)(N & (1<<pos));}
//==================== default ends ====================//
int arr[] = {111,222,333,444,555,666,777,888,999};
int find_num(int n){
	int mid = 9/2, start = 0, ed = 9;
	while(ed-start > 1){
		mid=(start+ed)>>1;
		if(arr[mid] > n){
			ed = mid;
		}else if(arr[mid] < n){
			start = mid;
		}else{
			return arr[mid];
		}
	}
	if(arr[start]==n)return arr[start];
	else return arr[start+1];
}
int main(){
	fastread
	int n;
	cin>>n;
	cout<<find_num(n)<<endl;
}