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>

const int XN=4e5+11,P=998244353;

void M(int &x) {
    ((x>=P) && (x-=P)) || ((x<0) && (x+=P));
}

int main() {
    std::ios::sync_with_stdio(0);
    std::cin.tie(0);
    //freopen("input","r",stdin);
    static char x[XN];
    int n;std::cin>>n>>(x+1);
    auto IsValid=[&](int i)->bool {
        return 2*n%i==0 && (2*n/i)%2==1;
    };
    static int f[XN],sum[XN];
    for(int i=1;i<=n;++i)
        sum[i]=(sum[i-1]*2LL+x[i]-'0')%P;
    for(int i=1;i<=2*n;++i) {
        if(IsValid(i)) {
            assert(i%2==0);
            f[i]=sum[i/2];
            bool tak=1;
            if(i!=2*n) {
                assert(i<=n);
                std::string s(x+1,x+1+i/2);
                while(s.length()!=n) {
                    s+=s.substr(s.length()-i/2,i/2);
                    for(int j=s.length()-i/2;j<s.length();++j)
                        s[j]^=1;
                }
                if(s>std::string(x+1,x+1+n))
                    tak=0;
            }
            M(f[i]+=tak);
        }
    }
    for(int i=1;i<=2*n;++i)
        if(IsValid(i))
            for(int j=2;i*j<=2*n;++j)
                if(IsValid(i*j))
                    M(f[i*j]-=f[i]);
    int Ans=0;
    for(int i=1;i<=2*n;++i)
        if(IsValid(i))
            M(Ans+=1LL*i*f[i]%P);
    
    int gs=P-1;
    for(int i=1;i<=2*n;++i)
        M(gs+=f[i]);
    assert(sum[n]==gs);
    std::cout<<Ans<<'\n';
    return 0;
}

