/*
s[i][j]表示i秒钟后第j个灯的情况
s[i][j] = (s[i-1][j] + s[i-1][j-1]) % 2;
令
Fn(1,len) = |s[i][0] s[i][1] …… s[i][len]|
F0(1,len)为输入状态
A(len len) = |1 1        |
             |  1 1      |
             |    ……   |
             |1         1|
Fn = F0 * A ^ n
这题中模板会超时，要把模板中Multiply()的%放到()外面才可以
*/
#include "Mat.h"
#include <iostream>
using namespace std;

int main()
{
    Mat A, F;
    mod = 2;
    char str[MAX];  
    int n, len, i;
    while (scanf("%d",&n)!=EOF)
    {
        scanf("%s",str); 
        len = strlen(str);
        F.ReSize(1, len);
        A.ReSize(len, len);
        for(i = 0; i < len; i++)
        {
            F.s[0][i] = str[i]-'0';
            A.s[i][i] = 1;
            if(i==0)
                A.s[len-1][i] = 1;
            else
                A.s[i-1][i] = 1;
        }
        A.Er_work(n);
        F.Multiply(A);
        for(i = 0; i < len; i++)
            printf("%d",F.s[0][i]); 
        printf("\n");
    }
    return 0;
}