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>
using namespace std;

const int maxn = 200010, P = 998244353;
int n, ans, f[maxn], g[maxn][2];
char s[maxn];

int main() {
    scanf("%d %s", &n, s + 1);
    g[0][0] = 1;
    for (int i = 1; i <= n; i++) if (!(n % i)) {
        if (!((n / i) & 1)) continue;
        for (int j = 1; j <= i; j++) {
            g[j][0] = g[j][1] = 0;
            for (int t = 0; t < 2; t++) {
                (g[j][1] += g[j - 1][1]) %= P;
                char c = t + '0';
                if (c <= s[j]) (g[j][c < s[j]] += g[j - 1][0]) %= P;
            }
        }
        bool flag = 1;
        for (int j = i + 1; j <= n; j++) {
            char c = s[j % i == 0 ? i : j % i];
            if (!(((j + i - 1) / i) & 1)) c ^= 1;
            if (c > s[j]) { flag = 0; break; }
            if (c < s[j]) break;
        }
        f[i] = g[i][1];
        if (flag) (f[i] += g[i][0]) %= P;
        for (int j = 1; j < i; j++) if (!(i % j)) {
            f[i] = (f[i] - f[j] + P) % P;
        }
        ans = (ans + 2LL * i * f[i]) % P;
    }
    printf("%d\n", ans);
    return 0;
}