Problem:
Score : 800 points

Problem Statement

Given are integers N and X. For each integer k between 0 and X (inclusive), find the answer to the following question, then compute the sum of all those answers, modulo 998244353.

Let us repeat the following operation on the integer k. Operation: if the integer is currently odd, subtract 1 from it and divide it by 2; otherwise, divide it by 2 and add 2^{N-1} to it. How many operations need to be performed until k returns to its original value? (The answer is considered to be 0 if k never returns to its original value.)

Constraints

1 \leq N \leq 2\times 10^5

0 \leq X < 2^N

X is given in binary and has exactly N digits. (In case X has less than N digits, it is given with leading zeroes.)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print the sum of the answers to the questions for the integers between 0 and X (inclusive), modulo 998244353.

Sample Input 1

3
111

Sample Output 1

40

For example, for k=3, the operation changes k as follows: 1,0,4,6,7,3. Therefore the answer for k=3 is 6.

Sample Input 2

6
110101

Sample Output 2

616

Sample Input 3

30
001110011011011101010111011100

Sample Output 3

549320998

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int
main(int argc, char *argv[])
{
    long long N, X;
    char strX[200000+1];
    long long try, all_try=0, base_try;
    long long tmp, pre_tmp, i;
    

    scanf("%lld", &N);
    scanf("%s", strX);

    X = 0;
    for (i=0; i<strlen(strX); i++) {
        X = X<<1;
        if (strX[i]=='1') {
            X |= 1;
        }
    }
//    printf("%lld\n", X);

    try = 0;
    tmp = 0;
    do {
        pre_tmp = tmp;
        if (tmp&1) {
            tmp = (tmp-1)>>1;
        }
        else {
            tmp = (tmp>>1) + (1LL<<(N-1));
        }
        try++;
//        printf("   %lld %lld\n", i, tmp);
    }
    while (tmp != 0);
    base_try = try;

    for (i=0; i<=X; i++) {
        try = 0;
        tmp = i;
        do {
            pre_tmp = tmp;
            if (tmp&1) {
                tmp = (tmp-1)>>1;
            }
            else {
                tmp = (tmp>>1) + (1LL<<(N-1));
            }
            try++;
//            printf("   %lld %lld\n", i, tmp);
        }
        while (tmp != i && pre_tmp!=tmp);
//        printf("%lld: %lld\n", i, try);
        if (pre_tmp!=tmp) {
            if (base_try!=try)
                break;
        }
    }
/*    
    printf("i: %lld: base %lld try %lld\n",i, base_try, try);
    printf("unit: %lld num_unit %lld(%lld %lld)\n", base_try*i+try, X/(i+1), X, i+1);
    printf("%lld\n", (base_try*i+try)*(X/(i+1)));
    printf("%lld\n", base_try*((X+1)-X/(i+1)*(i+1)));
*/
    all_try = (base_try * i + try)*(X/(i+1))+base_try*((X+1)-X/(i+1)*(i+1));

    printf("%lld\n", all_try%998244353);

    return 0;
}