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 <bits/stdc++.h>

const int moder = 998244353;
const int N = 200010;

char s[N];
int ans[N];

int main(){
    int n;
    scanf("%d%s", &n, s);
    std::vector <int> fact;
    for (int i = 1; i <= n; ++ i){
        if (n % i){
            continue;
        }
        if (n / i % 2 == 0){
            continue;
        }
        fact.push_back(i);
        int sum = 0;
        for (int j = 0; j < i; ++ j){
            sum = 2 * sum + s[j] - '0';
            sum -= sum >= moder ? moder : 0;
        }
        int sig = 0;
        for (int j = 0; j < n; ++ j){
            int real = (s[j % i] - '0') ^ ((j / i) & 1);
            if (sig == 0 && s[j] - '0' != real){
                sig = real > s[j] - '0' ? 1 : -1;
            }
        }
        if (sig <= 0){
            ++ sum;
            sum -= sum >= moder ? moder : 0;
        }
        ans[i] = sum;
    }
    for (auto u : fact){
        for (auto v : fact){
            if (v >= u){
                break;
            }
            if (u % v == 0){
                ans[u] -= ans[v];
                ans[u] += ans[u] < 0 ? moder : 0;
            }
        }
    }
    int ret = 0;
    for (int i = 1; i <= n; ++ i){
        ret = (ret + 2ll * i * ans[i]) % moder;
    }
    printf("%d\n", ret);
    return 0;
}
