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>
#define MN 400000
#define mod 998244353
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
char s[MN+5];
int n,ans,num,tot,res[MN+5];
int Solve(int x)
{
    int res=0;
    for(int i=1;i<=x;++i) res=(res*2+s[i]-'0')%mod;
    for(int i=1;i<=n;++i) 
    {
        int j=(i-1)%x+1,k=(s[j]-'0')^(1&((i-1)/x));
        if(k>s[i]-'0') return res;
        else if(k<s[i]-'0') return res+1;
    }
    return res+1;
}
int main()
{
    n=read();scanf("%s",s+1);
    for(int i=1;i<=n;++i) tot=(2LL*tot+s[i]-'0')%mod;
    tot=(tot+1)%mod;
    for(int i=1;i<=n;++i) if((n%i==0&&((n/i)&1))) res[i]=Solve(i);
    for(int i=1;i<=n;++i) if((n%i==0&&((n/i)&1))) for(int j=i<<1;j<=n;j+=i) res[j]=(res[j]-res[i]+mod)%mod;
    for(int i=1;i<=n;++i) if((n%i==0&&((n/i)&1))&&res[i])
    {
        ans=(ans+2LL*i*res[i])%mod;
        tot=(tot-res[i]+mod)%mod;
    }
    cout<<(ans+2LL*n*tot)%mod;
    return 0;
}