A Tide of Riverscape CodeForces - 989B （水题）

B. A Tide of Riverscape

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records.

The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.

In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1\le i\le \left|s\right|-p$, the $i$-th and $(i+p)$-th characters of $s$ are the same. Here $\left|s\right|$ is the length of $s$.

Input

The first line contains two space-separated integers $n$ and $p$ ($1\le p\le n\le 2000$) — the length of the given string and the supposed period, respectively.

The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.

Output

Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

Examples

input

Copy

10 7
1.0.1.0.1.

output

Copy

1000100010

input

Copy

10 6
1.0.1.1000

output

Copy

1001101000

input

Copy

10 9
1........1

output

Copy

No

Note

In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.

In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.

In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

题意：给出一个长为n且只有‘0’，‘1’，‘ .’ 的串，判断是否存在两个间隔为k的字符不相等（  字符‘ . ’可以任意变换成 ‘ 0 ’ 或 ‘  1 ’）

思路：emmmm  for 循环，直接讨论 i 和  i+k 个字符即可，如果i 和 i+k 不同，那么必定是满足条件的，如果i 和 i+k 相同，那么只有当 i  和 i+k 都为 ‘ . ' 才成立。

#include "iostream"
#include "string"
using namespace std;
int main()
{

    ios::sync_with_stdio(false);
    int n,k,flag=0;
    string str;
    cin>>n>>k>>str;
    for(int i=0;i<n;i++){
        if(i+k>=n) break;
        if(str[i]=='.'&&str[i+k]=='.'){str[i]='0',str[i+k]='1';flag=1;break;}
        if(str[i]!=str[i+k]){
            flag=1;
            if(str[i]!='.'&&str[i+k]!='.') break;
            if(str[i]=='0') str[i+k]='1';//依次讨论第 i 个字符的情况
            else if(str[i]=='1') str[i+k]='0';
            else if(str[i]=='.'){
                if(str[i+k]=='1') str[i]='0';
                else if(str[i+k]=='0') str[i]='1';
            }
            break;
        }
    }
    if(flag){
        for(int i=0;i<str.size();i++)
            if(str[i]!='.') cout<<str[i];//只要有一个满足即可，其他’.‘ 都用’1‘或着’0‘代替即可
            else cout<<"0";
    }
    else cout<<"No";
    cout<<endl;
    return 0;
}