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;

typedef long long ll;

const int mod = 998244353;
const int Maxn = 400005;

int pw2[Maxn];
int n;
char str[Maxn];
char my[Maxn];
int ans[Maxn];
int res;

int main()
{
    pw2[0] = 1;
    for (int i = 1; i < Maxn; i++)
        pw2[i] = 2ll * pw2[i - 1] % mod;
    scanf("%d", &n);
    scanf("%s", str);
    int cur = 0;
    for (int i = 0; i < n; i++)
        cur = (2ll * cur + int(str[i] - '0')) % mod;
    res = ll(cur + 1) * 2ll * n % mod;
    for (int i = n - 1; i >= 1; i--) if (n % i == 0 && n / i % 2) {
        int lft = i;
        for (int j = 0; j < i; j++) {
            lft--;
            if (str[j] == '1') ans[i] = (ans[i] + pw2[lft]) % mod;
            char A = str[j];
            char B = A == '1'? '0': '1';
            for (int z = j; z < n; z += i) {
                my[z] = A;
                swap(A, B);
            }
        }
        bool ok = true;
        for (int j = 0; j < n; j++)
            if (my[j] != str[j]) { ok = my[j] < str[j]; break; }
        if (ok) ans[i] = (ans[i] + 1) % mod;
    }
    for (int i = 1; i < n; i++) if (n % i == 0 && n / i % 2) {
        res = (res - ll(ans[i]) * (2 * n - 2 * i) % mod + mod) % mod;
        for (int j = i + i; j < n; j += i)
            ans[j] = (ans[j] - ans[i] + mod) % mod;
    }
    printf("%d\n", res);
    return 0;
}
