Lucky Numbers (easy) CodeForces - 96B

Petya loves lucky numbers. Everybody knows that positive integers are lucky if their decimal representation doesn't contain digits other than 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Lucky number is super lucky if it's decimal representation contains equal amount of digits 4 and 7. For example, numbers 47, 7744, 474477 are super lucky and 4, 744, 467 are not.

One day Petya came across a positive integer n. Help him to find the least super lucky number which is not less than n.

Input

The only line contains a positive integer n (1 ≤ n ≤ 10⁹). This number doesn't have leading zeroes.

Output

Output the least super lucky number that is more than or equal to n.

Please, do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Example

Input

4500

Output

4747

Input

47

Output

47

#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;

const long long int INF=10000000000000;

ll n,ans,mcnt;
char vis[12][2];
int a[12][2]={{4,7},{4,7},{4,7},{4,7},{4,7},{4,7},{4,7},{4,7},{4,7},{4,7},{4,7},{4,7}};

               
void DFS(ll p,ll q,int cnt,int x,int y)
{   ll tem=p*10+q;
    if(tem>=n&&x==y) ans=min(ans,tem);
    if(cnt==mcnt||cnt>12) return;
    
    for(int i=0;i<2;i++){
        if(vis[cnt][i]) continue;
        vis[cnt][i]=true;
        
        if(a[cnt][i]==4) DFS(tem,a[cnt][i],cnt+1,x+1,y);
        if(a[cnt][i]==7) DFS(tem,a[cnt][i],cnt+1,x,y+1);
         vis[cnt][i]=false;
    }
}

int main()
{    while(cin>>n){
        memset(vis,false,sizeof(vis));
        mcnt=0,ans=INF;
        int nn=n;
        while(nn){
            nn=nn/10;
            mcnt++;
        }
        if(mcnt%2==0) mcnt+=2;
        if(mcnt%2!=0) mcnt+=1;
        DFS(0,0,0,0,0);
        cout<<ans<<endl;
    }     
    return 0;
}