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.

Examples

Input

3
111


Output

40


Input

6
110101


Output

616


Input

30
001110011011011101010111011100


Output

549320998
#include <cstdio>

using namespace std;

const int mod = 998244353;
int n;
char s[200001];
long long a[200001];

long long calc(int c) {
    int i, j;
    long long ans = 0;
    
    for (i = 0; i < c; i++) {
        ans *= 2;
        if (ans >= mod) ans -= mod;
        
        if (s[i] == '1') {
            ans++;
            if (ans >= mod) ans -= mod;
        }
    }
    
    for (i = 0; i < n / c; i++) {
        for (j = 0; j < c; j++) {
            int x, y;
            
            if (i % 2 == 0) {
                x = s[i * c + j] - '0';
                y = s[j] - '0';
            } else {
                x = s[i * c + j] - '0';
                y = 1 - (s[j] - '0');
            }
            
            if (x > y) {
                return ans + 1;
            } else if (x < y) {
                return ans;
            }
        }
    }
    
    return ans + 1;
}

int main() {
    int i, j;
    long long ans = 0;
    
    scanf("%d", &n);
    scanf("%s", s);
    
    for (i = 1; i <= n; i++) {
        if (n % i == 0 && n / i % 2 == 1) {
            a[i] += calc(i);
            if (a[i] >= mod) a[i] -= mod;
            
            ans += a[i] * i * 2 % mod;
            
            for (j = i * 2; j <= n; j += i) {
                a[j] -= a[i];
                if (a[j] < 0) a[j] += mod;
            }
        }
    }
    
    printf("%lld\n", ans % mod);
    
    return 0;
}
