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 MAXN 400005
#define INF 1000000000
#define MOD 998244353
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
int n;
string str;
void add(int &a,int b) {a+=b; if(a>=MOD) a-=MOD;}
void dec(int &a,int b) {a-=b; if(a<0) a+=MOD;}
int res[MAXN];
int get_val(string str)
{
    int x=0;
    for(int i=0;i<(int)str.size();i++)
    {
        x=2LL*x%MOD;
        if(str[i]=='1') add(x,1);
    }
    return x+1;
}
bool check(string x)
{
    int m=x.size();
    string rx=x; for(int i=0;i<m;i++) rx[i]=('0'+'1'-rx[i]);
    string ss="";
    string tmp=str; for(int i=0;i<n;i++) tmp[i]=('0'+'1'-tmp[i]);
    for(int i=0;i<(2*n/m);i++)
    {
        if(i&1) ss+=rx;
        else ss+=x;
    }
    for(int i=0;i<(2*n)%m;i++)
    {
        if((n/m)&1) ss+=rx[i];
        else ss+=x[i];
    }
    return ss<=(str+tmp);
}
int get_ans(int x)
{
    if((2*n/x)%2==0) return 0;
    string t=str.substr(0,x/2);
    int ret=get_val(t);
    if(!check(t)) dec(ret,1);
    return ret;
}
int main()
{
    scanf("%d",&n);
    cin>>str;
    int ans=0;
    for(int i=1;i*i<=2*n;i++)
    {
        if(2*n%i) continue;
        res[i]=get_ans(i);
        res[2*n/i]=get_ans(2*n/i);
    }
    for(int i=1;i<=2*n;i++)
    {
        add(ans,1LL*i*res[i]%MOD);
        for(int j=2*i;j<=2*n;j+=i) if(res[j]) dec(res[j],res[i]);
    }
    printf("%d\n",ans);
    return 0;
}
