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 <iostream>
#include <vector>
#include <string>

using namespace std;

const int MOD = 998244353;

int main(){
    int N; cin >> N;
    string X; cin >> X;
    vector<long long> div;
    for(int i=1;i<=N;i++) if(N%i == 0 && N/i%2 == 1) div.push_back(i);
    vector<long long> cnt(N+1, 0);
    long long res = 0;
    for(auto& d : div){
        string s[2] = {"", ""};
        long long num = 0;
        for(int i=0;i<d;i++){
            num = (2 * num + X[i] - '0') % MOD;
            s[0] += X[i];
            s[1] += X[i] == '0' ? '1' : '0';
        }
        bool ok = true;
        for(int i=0;i<N;i++){
            if(X[i] < s[i/d%2][i%d]) { ok = false; break; }
            if(X[i] > s[i/d%2][i%d]) break;
        }
        cnt[d] = (num + (ok ? 1 : 0)) % MOD;
        for(auto& d2 : div){
            if(d2 >= d) break;
            if(d%d2 == 0) cnt[d] = (cnt[d] + MOD - cnt[d2]) % MOD;
        }
        res = (res + cnt[d] * 2 * d) % MOD;
    }
    cout << res << endl;
}