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 MAXNUM 211111
typedef long long ll;
#define rep(i,s,t) for(int i=s;i<t;i++)
char s[MAXNUM],st[MAXNUM];
ll num[MAXNUM];
vector<int> tmp;
const ll mod=998244353;
void add(ll &a,ll b)
{
    a+=b;
    if(a>=mod)a-=mod;
}
void sub(ll &a,ll b)
{
    a-=b;
    if(a<0)a+=mod;
}
int main()
{
    int n;scanf("%d%s",&n,s);
    rep(i,1,n+1)if((n%i==0)&&((n/i)&1))
        tmp.push_back(i);
    rep(i,0,tmp.size())
    {
        int k=tmp[i];
        ll r=0;
        rep(j,0,k)add(r,r),add(r,s[j]-'0');
        int x=0;
        rep(j,0,n/k)
        rep(kk,0,k)
        {
            st[x]= (j & 1) != 0 == !(s[kk] - '0');
            x++;
        }
        int flag=1;
        rep(kk,0,n)
        {
            if(s[kk]<st[kk]+'0'){flag=0;break;}
            else if(s[kk]>st[kk]+'0')break;
        }
        add(r,flag);num[i]=r;
    }
    ll res=0;
    rep(i,0,tmp.size())
    {
        rep(j, 0, i)
            if (tmp[i] % tmp[j] == 0)sub(num[i], num[j]);
        add(res, num[i] * 2 * tmp[i] % mod);
    }
    printf("%lld\n",res);
}