E. Binary Numbers AND Sum

time limit per test：1 second

memory limit per test：256 megabytes

input：standard input

output：standard output

You are given two huge binary integer numbers aa and bb of lengths nn and mm respectively. You will repeat the following process: if b>0 , then add to the answer the value a & band divide b by 2 rounding down (i.e. remove the last digit of bb ), 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=10102 (1010) and b=10002 (810) , then the value a & bwill be equal to 8 , not to 1000.

Input

The first line of the input contains two integers n and m (1≤n,m≤200000 ) — the length of aa and the length of bb correspondingly.

The second line of the input contains one huge integer aa . It is guaranteed that this number consists of exactly nn 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 mm 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

Note

The algorithm for the first example:

add to the answer 10102 & 11012=10002=810and set b:=110 ;
add to the answer 10102 & 1102=102=210 and set b:=11 ;
add to the answer 10102 & 112=102=210and set b:=1 ;
add to the answer 10102 & 12=02=010 and set b:=0 .

So the answer is 8+2+2+0=128+2+2+0=12 .

The algorithm for the second example:

add to the answer 10012 & 101012=12=110and set b:=1010b:=1010 ;
add to the answer 10012 & 10102=10002=810and set b:=101;
add to the answer 10012 & 1012=12=110and set b:=10 ;
add to the answer 10012 & 102=02=010and set b:=1;
add to the answer 10012 & 12=12=110and set b:=0.

So the answer is 1+8+1+0+1=111+8+1+0+1=11 .

题意：给定另个特别大的二进制数a,b，每次将a&b加入到结果中，并且将b除以二，直到b==0;求该结果；

解法：首先除以二操作相当于去掉二进制数b的最后一位数；其次如果每除以二后直接算a&b时间明显不够，如果开始将a,b的最后一位对齐放置，将b除以二相当于将b右移一位，超出的位忽略掉，那么一共右移m次，在这个过程中，a中的第i位（从左至右数）与b中的前max(0,m-(n-i))个数依次对齐一次，所以如果a[i]==1，计算b的前max(m-(n-i))个数中有多少个1就可以了（前缀和维护），然后乘以1<<(n-i),注意结果% 998244353；

附代码：

#include<bits/stdc++.h>

using namespace std;

#define pii pair<int, int>
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define per(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i>=b;i--)
#define all(x) x.begin(),x.end()
#define PER(i,x) for(auto i=x.begin();i!=x.end();i++)
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
typedef long long ll;
const double eps=1.0e-5;
const int maxn=200000+10;
const long long mod=998244353;

int a[maxn],b[maxn],sum[maxn],n,m;
char s[maxn];

int main()
{
	scanf("%d%d",&n,&m);
	scanf("%s",s+1);
	per(i,1,n) a[i]=(int)(s[i]-'0');
	scanf("%s",s+1);
	per(i,1,m) b[i]=(int)(s[i]-'0');
	sum[0]=0;
	per(i,1,m) sum[i]=sum[i-1]+b[i];
	
	int cur=1;
	ll ans=0;
	rep(i,n,1){
		if(a[i]==1) ans=(ans+((ll)sum[max(0,m-n+i)]*cur%mod))%mod;
		cur=(2*cur)%mod;
	}
	
	printf("%lld\n",ans);
}