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 Maxn 200007
#define modp 998244353
int n;
int a[Maxn],now[Maxn];
pair<int,int> que[Maxn];
int cnt=0;
int val[Maxn];
void solve(int len)
{
    ++cnt;
    que[cnt].first=len;
    que[cnt].second=0;
    for (int i=1;i<=len;i++)
        que[cnt].second=(2LL*que[cnt].second+a[i])%modp;
    bool check=true;
    for (int i=1;i<=len;i++)
        now[i]=a[i];
    for (int i=len+1;i<=n;i++)
        now[i]=now[i-len]^1;
    for (int i=1;i<=n;i++)
        if (a[i]!=now[i])
        {
            if (now[i]>a[i]) check=false;
            break;
        }
    if (check) que[cnt].second=(que[cnt].second+1)%modp;
}
int main()
{
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
    {
        char ch=getchar();
        while (ch!='0'&&ch!='1') ch=getchar();
        a[i]=ch-'0';
    }
    for (int i=1;i<=n;i++)
        if ((n%i==0)&&((n/i)%2==1)) solve(i);
    sort(que+1,que+cnt+1);
    int ans=0;
    for (int i=cnt;i;i--)
    {
        int tmp=(que[i].first-val[i]+modp)%modp;
        ans=(ans+1LL*tmp*que[i].second)%modp;
        for (int j=1;j<=cnt;j++)
            if (que[i].first%que[j].first==0) val[j]=(val[j]+tmp)%modp;
    }
    ans=(2LL*ans)%modp;
    printf("%d\n",ans);
    return 0;
}