As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the trains can get out in an order O2.

InputThe input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample Input.
OutputThe output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.
Sample Input

3 123 321
3 123 312

Sample Output

Yes.
in
in
in
out
out
out
FINISH
No.
FINISH
 
For the first Sample Input, we let train 1 get in, then train 2 and train 3.
So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.
In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.
Now we can let train 3 leave.
But after that we can't let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.
So we output "No.".

Hint

Hint

题目的大意： O1指的是进栈顺序，判断是否能以O2的方式输出;
分析：，进栈是可以边进边出栈.所以没进入一个元素就要判断当前元素是否满足出栈条件，判断的时候要用while循环。直到栈栈顶元素不满足出栈条件为止，
还有一点就是路径的记录， 可以开一个数组或者队列，进栈放1，出栈放0 当 输出yes的时候可以输出路径

AC代码：

 1 #include<iostream>
 2 #include<stack>
 3 #include<queue>
 4 #include<string>
 5 using namespace std;
 6 int main()
 7 {
 8     int n;
 9     while(cin>>n){
10         queue<int >que;
11         string a,b;
12         cin>>a>>b;
13         stack<int >s;
14         int pos=0;
15         for(int i=0;i<n;i++)
16         {
17             s.push(a[i]);
18             que.push(1);
19             while(s.size()&&s.top()==b[pos])
20             {
21                 s.pop();
22                 que.push(0);
23                 pos++;
24             }
25         }
26         if(pos==n) 
27         {
28             printf("Yes.\n");
29             while(que.size())
30             {
31                 if(que.front()==1)
32                 {
33                     printf("in\n");
34                     que.pop();
35                 }
36                 else {
37                     printf("out\n");
38                     que.pop();
39                 }
40             }
41             printf("FINISH\n");
42         }
43         else {
44             printf("No.\n");
45             printf("FINISH\n");
46         }
47     }
48     return 0;
49 }