F

Description

In a k bit 2's complement number, where the bits are indexed from 0 to k-1, the weight of the most significant bit (i.e., in position k-1), is -2^(k-1), and the weight of a bit in any position i (0 ≤ i < k-1) is 2^i. For example, a 3 bit number 101 is -2^2 + 0 + 2^0 = -3. A negatively weighted bit is called a negabit (such as the most significant bit in a 2's complement number), and a positively weighted bit is called a posibit.         A Fun number system is a positional binary number system, where each bit can be either a negabit, or a posibit. For example consider a 3-bit fun number system Fun3, where bits in positions 0, and 2 are posibits, and the bit in position 1 is a negabit. (110)Fun3 is evaluated as 2^2-2^1 + 0 = 3. Now you are going to have fun with the Fun number systems! You are given the description of a k-bit Fun number system Funk, and an integer N (possibly negative. You should determine the k bits of a representation of N in Funk, or report that it is not possible to represent the given N in the given Funk. For example, a representation of -1 in the Fun3 number system (defined above), is 011 (evaluated as 0 - 2^1 + 2^0), and representing 6 in Fun3 is impossible.      

Input

The first line of the input file contains a single integer t (1 ≤ t ≤ 10), the number of test cases, followed by the input data for each test case. Each test case is given in three consecutive lines. In the first line there is a positive integer k (1 ≤ k ≤ 64). In the second line of a test data there is a string of length k, composed only of letters n, and p, describing the Fun number system for that test data, where each n (p) indicates that the bit in that position is a negabit (posibit).         The third line of each test data contains an integer N (-2^63 ≤ N < 2^63), the number to be represented in the Funk number system by your program.      

Output

For each test data, you should print one line containing either a k-bit string representing the given number N in the Funk number system, or the word Impossible, when it is impossible to represent the given number.      

Sample Input

2
3
pnp
6
4
ppnn
10

Sample Output

Impossible
1110




这个题~  我这么笨 一定想不到简单的方法！！！  只有各种复杂办法  基本是不可行的
所以 必然需要求助！！

思路
1.n为奇数    最后一位一定是1，因为只有2的0次方可产生奇数，其他都为偶数。
       最后一位若为p（正）  则n=（n-1)/2
              若为n（负）  则n=（n+1)/2  
  n为偶数   最后一位一定是0，n=n/2   
2.接下来判断倒数第二位  将其重复步骤1，直至判断完这个字符串系统 
  最后n若为0  则可表示出
  n不为0  则结果为impossible


代码是这样

#include<iostream>
#include<string>
using namespace std;
int main() {
    int t,k;
   
    cin>>t;
    while(t--){ 
        __int64 n;
        string str;
        cin>>k>>str>>n;
        int j=0;
        int *p=new int[k];
        for(int i=k-1;i>=0;i--){
            if(n%2==1||n%2==-1){
                if(str[i]=='p')n=(n-1)/2;
                else n=(n+1)/2;
                p[j++]=1;
            }
            else {
                n/=2;
                p[j++]=0;
            }
        }
        if(n)cout<<"Impossible"<<endl;
        else {
            for(int i=j-1;i>=0;i--)cout<<p[i];
            cout<<endl;
        }
        delete []p;
    }
    //system("pause");
    return 0;
}