Binary Numbers AND Sum CodeForces

You are given two huge binary integer numbers $a$

The value $a & b$

Note that you should add the value $a & b$

Input

The first line of the input contains two integers $n$

The second line of the input contains one huge integer $a$

The third line of the input contains one huge integer $b$

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 $1010_{2} & 1101_{2} = 1000_{2} = 8_{10}$
add to the answer $1010_{2} & 110_{2} = 10_{2} = 2_{10}$
add to the answer $1010_{2} & 11_{2} = 10_{2} = 2_{10}$
add to the answer $1010_{2} & 1_{2} = 0_{2} = 0_{10}$

So the answer is $8 + 2 + 2 + 0 = 12$

The algorithm for the second example:

add to the answer $1001_{2} & 10101_{2} = 1_{2} = 1_{10}$
add to the answer $1001_{2} & 1010_{2} = 1000_{2} = 8_{10}$
add to the answer $1001_{2} & 101_{2} = 1_{2} = 1_{10}$
add to the answer $1001_{2} & 10_{2} = 0_{2} = 0_{10}$
add to the answer $1001_{2} & 1_{2} = 1_{2} = 1_{10}$

So the answer is $1 + 8 + 1 + 0 + 1 = 11$

$1 + 8 + 1 + 0 + 1 = 11$

4 4
1010
1101

我们看b，1101 从最后一位看起，最后一位的1，只能&上一次a的最后一位，然后就被消除掉了（右移）
而倒数第二位的0，无论&上多少个数（不管是0还是1） 都不会对答案做出贡献。
继续看倒数第三位的1，他会&上a中的后三位才会被消除掉，会&上a中倒数第2位的1，那么会对答案产生2的贡献。
再看b的第一位的1，他会&上a中的后四位才会被消除掉，&上a中倒数第2个的1和倒数第4位的1，会对答案产生8+2的贡献，
加起来答案就是12。
不知道大家有没有发现什么规律？
b中每一位1都会&上a中对应位置以及后面的所有位。并且b这一位的贡献值就是a中这一位以及之后的数组成的二进制数的十进制数值大小。
那么我们不妨对a进行通过快速幂取模来求出a中每一位数值的前缀和，然后b中每一位1直接去加上那么数值就是贡献。
又因为a和b可能不一样长，我们为了方便把a和b翻转后再计算。
细节见代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define ALL(x) (x).begin(), (x).end()
#define rt return
#define dll(x) scanf("%I64d",&x)
#define xll(x) printf("%I64d
",x)
#define sz(a) int(a.size())
#define all(a) a.begin(), a.end()
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define eps 1e-6
#define gg(x) getInt(&x)
#define db(x) cout<<"== [ "<<x<<" ] =="<<endl;
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/
//string a,b;
int n,m;
char a[maxn];
char b[maxn];
const ll mod=998244353ll;
ll sum[maxn];
int main()
{
    //freopen("D:\common_text\code_stream\in.txt","r",stdin);
    //freopen("D:\common_text\code_stream\out.txt","w",stdout);
    gbtb;
    cin>>n>>m;
    cin>>a>>b;
    ll x=m-n;
    reverse(a,a+n);
    reverse(b,b+m);

    if(a[0]=='1')
    {
        sum[0]=1ll;
    }
    for(ll i=1ll;i<max(n,m);++i)
    {
        if(a[i]=='1')
        {
            sum[i]=sum[i-1]+powmod(2ll,i,mod);
            sum[i]=(sum[i]+mod)%mod;
        }else
        {
            sum[i]=sum[i-1];
        }
    }
    ll ans=0ll;
    for(ll i=0ll;i<max(n,m);++i)
    {
        if(b[i]=='1')
        {
            ans+=sum[i];
            ans=(ans+mod)%mod;
        }
    }
    cout<<ans<<endl;



    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '
');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

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