Codeforces Round #609 (Div. 2) C.Long Beautiful Integer

Codeforces Round #609 (Div. 2) C.Long Beautiful Integer

You are given an integer x of n digits a1,a2,…,an, which make up its decimal notation in order from left to right.

Also, you are given a positive integer k<n.

Let’s call integer b1,b2,…,bm beautiful if bi=bi+k for each i, such that 1≤i≤m−k.

You need to find the smallest beautiful integer y, such that y≥x.

Input

The first line of input contains two integers n,k (2≤n≤200000,1≤k<n): the number of digits in x and k.

The next line of input contains n digits a1,a2,…,an (a1≠0, 0≤ai≤9): digits of x.

Output

In the first line print one integer m: the number of digits in y.

In the next line print m digits b1,b2,…,bm (b1≠0, 0≤bi≤9): digits of y.

Examples

input

3 2
353

output

3
353

input

4 2
1234

output

4
1313

比赛时调了半天，看了题解发现好简单……，就调整前k个数即可，找到k个数中最后一个不为9的数字的位置，把它加一，后面的9全部置0即可：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

int main(){
    string s1,s2="";
    int n,k;
    cin>>n>>k>>s1;
    for(int i=0;i<n;i++){
        if(i<k) s2+=s1[i];
        else s2+=s2[i-k];
    }
    if(s2>=s1) cout<<n<<endl<<s2<<endl;
    else{
        int u=0;
        for(int i=k-1;i>=0;i--){
            if(s1[i]!='9') {s1[i]=s1[i]+1;u=i;break;}
        }
        for(int i=u+1;i<k;i++){
            s1[i]='0';
        }
        string ans="";
        for(int i=0;i<n;i++){
            if(i<k) ans+=s1[i];
            else ans+=ans[i-k];
        }
        cout<<n<<endl<<ans<<endl;
    }
}