[codeforces1066E]Binary Numbers AND Sum

time limit per test : 1 second
memory limit per test : 256 megabytes

You are given two huge binary integer numbers aand b of lengths n and m respectively. You will repeat the following process: if $b>0$ , then add to the answer the value $a$ & $b$ and divide $b$ by $2$ rounding down (i.e. remove the last digit of $b$ ), and repeat the process again, otherwise stop the process.The value $a$ & $b$ means bitwise AND of $a$ and $b$ . Your task is to calculate the answer modulo $998244353$

Note that you should add the value $a$ & $b$ to the answer in decimal notation, not in binary. So your task is to calculate the answer in decimal notation. For example, if $a=1010_2 (10_{10})$ and $b=1000_2 (8_{10})$ , then the value $a$ & $b$ will be equal to $8$ , not to $1000$

Input

The first line of the input contains two integers $n$ and $m (1≤n,m≤2⋅10^5)$ — the length of $a$ and the length of $b$ correspondingly.
The second line of the input contains one huge integer $a$ . It is guaranteed that this number consists of exactly n zeroes and ones and the first digit is always 1
The third line of the input contains one huge integer b. It is guaranteed that this number consists of exactly m zeroes and ones and the first digit is always 1

Output

Print the answer to this problem in decimal notation modulo $998244353$

Examples

Input

4 4
1010
1101

Output

Input

4 5
1001
10101

Output

题意：
给两个二进制数 $a$ , $b$ , 位数分别为 $n$ , $m (1≤n,m≤2⋅10^5)$ ，有下列操作：
step1. $x:=$ $a$ & $b$
step2. $b:=b/2$
step3. 如果 $b>0$ 回到step1
求所有 $x$ 的总和，由于总和可能会很大，所以总和对 $998244353$ 取模

题解：
对于 $a$ 来说，他处于从后往前数第 $i$ 位上的 $1$ 对答案的贡献，是 $1<<i$ ，记作 $ret_i$ 。
对于 $b$ 来说，我们单独考虑他每个 $1$ 对答案的贡献，是这个 $1$ 所处的位置在 $a$ 中所有能被他 $AND$ 到的 $1$ 的贡献之和,这个对 $ret_i$ 处理一个后缀和就行了。
$cop_i$ 记作从第n位到第i位所有的 $ret_i$ 之和

#include<bits/stdc++.h>
#define LiangJiaJun main
#define MOD 998244353
using namespace std;
int bin[200004],n,m;
char a[200004],b[200004];
int cop[200004];
int w33ha(){
    scanf("%s",a+1);
    scanf("%s",b+1);
    cop[n+1]=0;
    for(int i=n;i>=1;i--){
        if(a[i]=='0')cop[i]=cop[i+1];
        else cop[i]=(cop[i+1]+bin[n-i])%MOD;
    }
    int ans=0;
    for(int i=1;i<=m;i++){
        if(b[i]=='0')continue;
        int bg=m-i+1;
        if(bg>n)bg=1;
        else bg=n-bg+1;
        ans=(ans+cop[bg])%MOD;
    }
    printf("%d\n",ans);
    return 0;
}
int LiangJiaJun(){
    bin[0]=1;
    for(int i=1;i<=200001;i++)bin[i]=(bin[i-1]<<1)%MOD;

    while(scanf("%d%d",&n,&m)!=EOF)w33ha();
    return 0;
}