CF989B A Tide of Riverscape 思维 第七题

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 ss 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 pp 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 pp is considered a period of string ss, if for all 1≤i≤|s|−p1≤i≤|s|−p, the ii-th and (i+p)(i+p)-th characters of ssare the same. Here |s||s| is the length of ss.

Input

The first line contains two space-separated integers nn and pp (1≤p≤n≤20001≤p≤n≤2000) — the length of the given string and the supposed period, respectively.

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

Output

Output one line — if it's possible that pp 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, 77 is not a period of the resulting string because the 11-st and 88-th characters of it are different.

In the second example, 66 is not a period of the resulting string because the 44-th and 1010-th characters of it are different.

In the third example, 99 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.

啊啊啊啊啊啊，简直了，佩服我自己，一个等于符号写成赋值符号，调半天bug

题意：给你n个字符串，问你每隔k个是否右不相等的，没有输出"No"，有输出一个符合不相等的（注意‘.'可以是0或者1，所以输出的时候'.'是会根据前后的数字发生变化的）

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e6 + ;
const int mod = 1e9 + ;
typedef long long ll;
int main(){
    std::ios::sync_with_stdio(false);
    ll n, k;
    while( cin >> n >> k ) {
        string s;
        cin >> s;
        ll num;
        bool flag = false;
        for( ll i = ; i < s.length(); i ++ ) {
            if( i + k < s.length() ) {
                if( s[i] != s[i+k] || ( s[i] == s[i+k] && s[i] == '.' ) ) {
                    if( s[i] == s[i+k] ) {
                        s[i] = '', s[i+k] = '';
                    } else {
                        if( s[i] == '' ) {
                            s[i+k] = '';
                        } else if( s[i] == '' ) {
                            s[i+k] = '';
                        } else {
                            if( s[i+k] == '' ) {  //这里等于号写成了赋值符号，调了半天bug，吐血ing
                                s[i] = '';
                            } else {
                                s[i] = '';
                            }
                        }
                    }
                    flag = true;
                    break;
                }
            }
        }
        if( flag ) {
            for( ll i = ; i < s.length(); i ++ ) {
                if( s[i] == '.' ) {
                    cout << "";
                } else {
                    cout << s[i];
                }
            }
        } else {
            cout << "No";
        }
        cout << endl;
    }
    return ;
}