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C - Division by Two with Something /

Time Limit: 2 sec / Memory Limit: 1024 MB

### 問題文

• 現在の整数が奇数なら、1 を引いて 2 で割る。そうでなければ、2 で割って 2^{N-1} を足す。

### 制約

• 1 \leq N \leq 2\times 10^5
• 0 \leq X < 2^N
• X2 進表記でちょうど N 桁与えられる (XN 桁より少ない場合、leading zero をつけて与えられる。)
• 入力はすべて整数である

### 入力

N
X


### 出力

0 以上 X 以下のすべての整数に対し、 操作を繰り返すことによってもとの整数に戻るまでの操作回数 (戻らない場合 0) を足し合わせた値を 998244353 で割ったあまりを出力せよ。

### 入力例 1

3
111


### 出力例 1

40


### 入力例 2

6
110101


### 出力例 2

616


### 入力例 3

30
001110011011011101010111011100


### 出力例 3

549320998


Score : 800 points

### Problem Statement

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.

### Sample Input 1

3
111


### Sample Output 1

40


For example, for k=3, the operation changes k as follows: 1,0,4,6,7,3. Therefore the answer for k=3 is 6.

### Sample Input 2

6
110101


### Sample Output 2

616


### Sample Input 3

30
001110011011011101010111011100


### Sample Output 3

549320998