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;

using LL = long long;

const int N = (int)2e5 + 5, mod = 998244353;
int n, nString[N << 1], pow2[N];
char x[N];

int add(int a, int b) {
    a += b;
    if (a >= mod) a -= mod;
    return a;
}
int sub(int a, int b) { return add(a, mod - b); }
int mul(int a, int b) { return (int)( (LL)a * b % mod); }

bool check(int per) {
    for (int i = 1; i <= n; ++i) {
        char cur;
        if (i % per) cur = x[i % per];
        else cur = x[per];

        if ( ( (i - 1) / per) & 1) {
            if (cur == '0') cur = '1';
            else cur = '0';
        }

        if (cur != x[i]) return cur < x[i];
    }
    return true;
}

void solve(int per) {
    per /= 2;
    for (int i = 1; i <= per; ++i) {
        if (x[i] == '1') nString[per << 1] = add(nString[per << 1], pow2[per - i]);
    }
    nString[per << 1] = add(nString[per << 1], check(per) );
}

int main() {
    scanf(" %d %s", &n, x + 1);

    pow2[0] = 1; for (int i = 1; i < N; ++i) pow2[i] = (int)(2 * pow2[i - 1] % mod);

    vector<int> divs;
    for (int i = 1; i <= (n << 1); ++i) if ( (n << 1) % i == 0 && ( (n << 1) / i) & 1) {
        solve(i);
        divs.emplace_back(i);
    }

    int ans = 0;
    for (auto per : divs) {
        for (auto per2 : divs) if (per != per2 && per % per2 == 0) nString[per] = sub(nString[per], nString[per2]);
    }
    for (auto per : divs) ans = add(ans, mul(nString[per], per) );

    printf("%d", ans);

    return 0;
}
