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;

#define N 200200

int n, up[N], pw[N], ans[N];
const int mod = 998244353;
char s[N];

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

vector <int> vec;

int main() {
    pw[0] = 1;
    for (int i = 1; i < N; i ++) pw[i] = (pw[i-1] << 1) % mod;
    scanf("%d %s", &n, s);
    for (int i = 0; i < n; i ++) s[i] -= '0';
    int d;
    for (d = 1; d * d < n; d ++) if (n % d == 0) {
        int dd = n / d;
        if (d & 1) vec.push_back(dd);
        if (dd & 1) vec.push_back(d);
    }
    if (d * d == n && d & 1) vec.push_back(d);
    int mn = n;
    for (auto d : vec) {
        int k;
        for (k = d; k < n; k ++) if (s[k] == s[k-d]) break;
        up[d] = k;
        if (k == n) mn = min(mn, d);
    }
    for (int i = 0; i < n; i ++) {
        if (s[i]) {
            for (auto d : vec) {
                if (d > i) add(ans[d], pw[d-i-1]);
                else if (up[d] == i) add(ans[d], 1);
            }
        }
    }

    int sz = vec.size(); sort(vec.begin(), vec.end());
    for (int i = 0; i < sz; i ++) {
        int dd = vec[i];
        for (int j = i + 1; j < sz; j ++) {
            int d = vec[j];
            if (d % dd == 0 && (d / dd) & 1) add(ans[d], mod - ans[dd]);
        }
    }
    int rlt = mn;
    for (auto d : vec) rlt = (rlt + 1ll * d * ans[d]) % mod;
    rlt <<= 1; if (rlt >= mod) rlt -= mod;
    printf("%d\n", rlt);
}
