C. Postcard time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output

Andrey received a postcard from Irina. It contained only the words "Hello, Andrey!", and a strange string consisting of lowercase Latin letters, snowflakes and candy canes. Andrey thought that this string is an encrypted message, and decided to decrypt it.

Andrey noticed that snowflakes and candy canes always stand after the letters, so he supposed that the message was encrypted as follows. Candy cane means that the letter before it can be removed, or can be left. A snowflake means that the letter before it can be removed, left, or repeated several times.

For example, consider the following string:

This string can encode the message «happynewyear». For this, candy canes and snowflakes should be used as follows:

• candy cane 1: remove the letter w,
• snowflake 1: repeat the letter p twice,
• candy cane 2: leave the letter n,
• snowflake 2: remove the letter w,
• snowflake 3: leave the letter e.

Please note that the same string can encode different messages. For example, the string above can encode «hayewyar», «happpppynewwwwwyear», and other messages.

Andrey knows that messages from Irina usually have a length of kk letters. Help him to find out if a given string can encode a message of kkletters, and if so, give an example of such a message.

Input

The first line contains the string received in the postcard. The string consists only of lowercase Latin letters, as well as the characters «*» and «?», meaning snowflake and candy cone, respectively. These characters can only appear immediately after the letter. The length of the string does not exceed 200200.

The second line contains an integer number kk (1≤k≤2001≤k≤200), the required message length.

Output

Print any message of length kk that the given string can encode, or «Impossible» if such a message does not exist.

Examples input Copy

hw?ap*yn?eww*ye*ar
12

output Copy

happynewyear

input Copy

ab?a
2

output Copy

aa

input Copy

ab?a
3

output Copy

aba

input Copy

ababb
5

output Copy

ababb

input Copy

ab?a
1

output Copy

Impossible

 

 

題意就是給你一個字串，其中？可以有兩種操作，1.將前面的字母放到當前位置，其實就是去掉？2.去掉前面的字母，其實就是去掉前面的字母去掉？

*的有三種操作，1.將前面的字母放到當前位置，就是去掉* 2.去掉前面的字母，就是去掉前面的字母和* 3.前面的字母可以重複任意次

假設是abc*，可以是abc，可以是ab，可以是abcccccc，就是這樣的，給你一個字串，問你能不能將字串變成要求的長度，最後的字串不能有？和*。

直接將純字母的長度先找出來，然後判斷再進行操作就可以了。

 

程式碼:

 1 //C
 2 #include<iostream>
 3 #include<cstring>
 4 #include<cstdio>
 5 #include<algorithm>
 6 #include<cmath>
 7 using namespace std;
 8 typedef long long ll;
 9 const int maxn=1e5+10;
10 
11 char s[200];
12 
13 int main()
14 {
15     scanf("%s",s);
16     int n;
17     cin>>n;
18     int len=strlen(s);
19     int candy=0,snow=0;
20     for(int i=0;i<len;i++){
21         if(s[i]=='?') candy++;
22         else if(s[i]=='*') snow++;
23     }
24     int l=len-candy-snow;
25     if(l==n){
26         for(int i=0;i<len;i++){
27             if(s[i]!='?'&&s[i]!='*') cout<<s[i];
28         }
29         cout<<endl;
30     }
31     else if(l<n&&snow>0){
32         int c;
33         for(int i=0;i<len;i++){
34             if(s[i]=='?') s[i]='#';
35             if(s[i]=='*'){
36                 if(l<n){
37                     c=n-l;
38                     s[i]='!';
39                     //cout<<s[i]<<endl;
40                     l=n;
41                 }
42                 else{
43                     s[i]='#';
44                 }
45             }
46         }
47         for(int i=0;i<len;i++){
48             if(s[i]!='#'){
49                 if(s[i]=='!'){
50                     for(int j=0;j<c;j++)
51                         cout<<s[i-1];
52                 }
53                 else cout<<s[i];
54             }
55         }
56         cout<<endl;
57     }
58     else if(l>n&&l-(candy+snow)<=n){
59         for(int i=0;i<len;i++){
60             if((s[i]=='?'||s[i]=='*')&&l>n){
61                 s[i-1]='#';
62                 s[i]='#';
63                 l--;
64             }
65             else if(s[i]=='?'||s[i]=='*')
66                 s[i]='#';
67         }
68         for(int i=0;i<len;i++){
69             if(s[i]!='#') cout<<s[i];
70         }
71         cout<<endl;
72     }
73     else cout<<"Impossible"<<endl;
74 }