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 cs const
#define pb push_back
using namespace std;
cs int Mod = 998244353;
int add(int a, int b){ return a + b >= Mod ? a + b - Mod : a + b; }
int dec(int a, int b){ return a - b < 0 ? a - b + Mod : a - b; }
int mul(int a, int b){ return 1ll * a * b % Mod; }
void Add(int &a, int b){ a = add(a,b); }
void Dec(int &a, int b){ a = dec(a,b); }
void Mul(int &a, int b){ a = mul(a,b); }
cs int N = 2e5 + 50;
int n, a[N], f[N]; char S[N];
void chk(int len){
    if(len & 1) return; 
    static int b[N]; int ans = 0;
    for(int i = 1; i <= len / 2; i++) 
    b[i] = a[i], ans = add(mul(ans, 2), a[i]);
    for(int i = len / 2 + 1; i <= n; i++)
    b[i] = b[i - len / 2] ^ 1; 
    bool ok = 1;
    for(int i = 1; i <= n && ok; i++)
        if(a[i] < b[i]) ok = 0;
        else if(a[i] > b[i]) break;
    f[len] = add(ans, ok);
}
int main(){
    #ifdef FSYolanda
    freopen("1.in","r",stdin);
    #endif
    scanf("%d%s", &n, S+1);
    for(int i = 1; i <= n; i++)
        a[i] = S[i] - '0';
    int t = n + n;
    for(int i = 1; i <= t; i++)
        if(t % i == 0 && n % i) chk(i); 
    for(int i = 1; i <= t; i++) if(f[i])
        for(int j = i + i; j <= t; j += i)
            Dec(f[j], f[i]);
    int ans = 0;
    for(int i = 1; i <= t; i++) 
    if(t % i == 0 && n % i)
    Add(ans, mul(f[i], i)); 
    cout << ans; return 0;
  
}