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 mod = 998244353;

inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int sub(int x, int y) { return x - y >= 0 ? x - y : x - y + mod; }

const int N = 2e5 + 5;

int n, m, ans, f[N], len[N];
char str[N];

int check(int len) {
    for (int i = 0, k = 0; i < n; i += len, k ^= 1) {
        for (int j = 0; j < len; j++) {
            if ((str[j] ^ k) == (str[i + j])) { continue; }
            if ((str[j] ^ k) < (str[i + j])) { return 1; }
            return 0;
        }
    } return 1;
}
int solve(int len) {
    int res = 0;
    for (int i = 0; i < len; i++) {
        res = add(add(res, res), str[i] - '0');
    } return res + check(len);
}

int main() {
    scanf("%d%s", &n, str);
    for (int i = 1; i <= n; i++) {
        if (n % i == 0 && n / i % 2 == 1) { len[m++] = i; }
    }
    for (int i = 0; i < m; i++) {
        f[i] = solve(len[i]);
        for (int j = 0; j < i; j++) {
            if (len[i] % len[j] == 0 && len[i] / len[j] % 2 == 1) { f[i] = sub(f[i], f[j]); }
        } ans = add(ans, 1ll * len[i] * 2 % mod * f[i] % mod);
    } printf("%d\n", ans);
    return 0;
}