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 ll long long

ll modpower(ll a, ll r, ll mod){ //a^r
    ll x = 1;
    a %= mod;
    while (r > 0) {
        if (r & 1) x = x * a % mod;
        r >>= 1;
        a = a * a % mod;
    }
    return x;
}

int main() {
    ll mod = 998244353;
    int N;
    string S;
    cin >> N;
    cin >> S;
    vector<ll> div;
    for (int i = 1; i <= N*2; i++) {
        if (2*N%i == 0 && N%i != 0) {
            div.push_back(i);
        }
    }
    map<ll, ll> mp;
    for (ll d : div) {
        ll x = 0;
        for (int i = 0; i < d/2; i++) {
            x *= 2;
            if (S[i] == '1') x += 1;
            x %= mod;
        }
        x++;
        bool check = true;
        for (int i = d/2; i < N; i++) {
            int period = i/(d/2);
            int digit = (int)(S[i%(d/2)] - '0');
            if (period%2 == 1) digit = 1-digit;
            if (digit < (int)(S[i]-'0')) {
                break;
            }else if (digit > (int)(S[i]-'0')) {
                check = false;
                break;
            }
        }
        if (!check) x--;
        mp[d] += x;
        for (int i = d*2; i <= N*2; i += d) {
            mp[i] += mod-mp[d];
            mp[i] %= mod;
        }
    }
    ll ans = 0;
    for (ll d : div) {
        ans += mp[d] * d % mod;
    }
    cout << (ans%mod+mod)%mod << endl;
    return 0;
}