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

23pnp64ppnn10

Sample Output

Impossible1110 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
       
        #include
        
         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"<
         
          =0;i--)cout<