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<cstdio>
#define RI register int
#define CI const int&
using namespace std;
const int N=400005,mod=998244353;
int n,ans,num[N],g[N],tot; char s[N],t[N];
inline void inc(int& x,CI y)
{
    if ((x+=y)>=mod) x-=mod;
}
inline void dec(int& x,CI y)
{
    if ((x-=y)<0) x+=mod;
}
int main()
{
    RI i,j,k; for (scanf("%d%s",&n,s+1),k=1;k<=(n<<1);++k)
    if ((n<<1)%k==0&&(((n<<1)/k)&1))
    {
        for (num[++tot]=k,i=1;i<=(k>>1);++i)
        g[tot]=((g[tot]<<1)+s[i]-48)%mod,t[i]=s[i];
        for (i=(k>>1)+1;i<=n;++i) t[i]=((t[i-(k>>1)]-48)^1)+48;
        for (++g[tot],i=1;i<=n;++i) if (t[i]!=s[i])
        { g[tot]-=(t[i]>s[i]); break; }
    }
    for (i=1;i<tot;++i) for (j=i+1;j<=tot;++j)
    if (num[j]%num[i]==0) dec(g[j],g[i]);
    for (i=1;i<=tot;++i) inc(ans,1LL*num[i]*g[i]%mod);
    return printf("%d",ans),0;
}