#include <bits/stdc++.h>
using namespace std;
long long n, m;
string s1, s2;
long long mod = 998244353;
long long power(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((a % mod * a % mod) % mod, b / 2);
  else
    return (a % mod * power((a % mod * a % mod) % mod, b / 2) % mod) % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  cin >> s1 >> s2;
  string temp = "";
  if (s1.size() < s2.size()) {
    int l = s2.size() - s1.size();
    while (l > 0) {
      temp += '0';
      l--;
    }
    temp += s1;
    s1 = temp;
    n = s1.size();
  } else {
    int l = s1.size() - s2.size();
    while (l > 0) {
      temp += '0';
      l--;
    }
    temp += s2;
    s2 = temp;
    m = s2.size();
  }
  long long prefix[n + 1];
  prefix[n] = 0;
  vector<long long> arr;
  for (int i = n - 1; i >= 0; i--) {
    if (s1[i] == '1') {
      arr.push_back(i);
      prefix[i] = (prefix[i + 1] % mod + power(2, (n - 1) - i) % mod) % mod;
    } else
      prefix[i] = prefix[i + 1];
  }
  long long sum = 0, k = 0;
  sort(arr.begin(), arr.end());
  for (int i = 0; i < m; i++) {
    if (s2[i] == '1') {
      sum = (sum % mod + prefix[i] % mod) % mod;
    }
  }
  cout << sum << endl;
}

### Prompt

Create a solution in Cpp for the following problem:

You are given two huge binary integer numbers a and 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

12


Input

4 5
1001
10101


Output

11

Note

The algorithm for the first example: 

  1. add to the answer 1010_2~ \&~ 1101_2 = 1000_2 = 8_{10} and set b := 110; 
  2. add to the answer 1010_2~ \&~ 110_2 = 10_2 = 2_{10} and set b := 11; 
  3. add to the answer 1010_2~ \&~ 11_2 = 10_2 = 2_{10} and set b := 1; 
  4. add to the answer 1010_2~ \&~ 1_2 = 0_2 = 0_{10} and set b := 0. 



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

The algorithm for the second example: 

  1. add to the answer 1001_2~ \&~ 10101_2 = 1_2 = 1_{10} and set b := 1010; 
  2. add to the answer 1001_2~ \&~ 1010_2 = 1000_2 = 8_{10} and set b := 101; 
  3. add to the answer 1001_2~ \&~ 101_2 = 1_2 = 1_{10} and set b := 10; 
  4. add to the answer 1001_2~ \&~ 10_2 = 0_2 = 0_{10} and set b := 1; 
  5. add to the answer 1001_2~ \&~ 1_2 = 1_2 = 1_{10} and set b := 0. 



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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m;
string s1, s2;
long long mod = 998244353;
long long power(long long a, long long b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0)
    return power((a % mod * a % mod) % mod, b / 2);
  else
    return (a % mod * power((a % mod * a % mod) % mod, b / 2) % mod) % mod;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  cin >> s1 >> s2;
  string temp = "";
  if (s1.size() < s2.size()) {
    int l = s2.size() - s1.size();
    while (l > 0) {
      temp += '0';
      l--;
    }
    temp += s1;
    s1 = temp;
    n = s1.size();
  } else {
    int l = s1.size() - s2.size();
    while (l > 0) {
      temp += '0';
      l--;
    }
    temp += s2;
    s2 = temp;
    m = s2.size();
  }
  long long prefix[n + 1];
  prefix[n] = 0;
  vector<long long> arr;
  for (int i = n - 1; i >= 0; i--) {
    if (s1[i] == '1') {
      arr.push_back(i);
      prefix[i] = (prefix[i + 1] % mod + power(2, (n - 1) - i) % mod) % mod;
    } else
      prefix[i] = prefix[i + 1];
  }
  long long sum = 0, k = 0;
  sort(arr.begin(), arr.end());
  for (int i = 0; i < m; i++) {
    if (s2[i] == '1') {
      sum = (sum % mod + prefix[i] % mod) % mod;
    }
  }
  cout << sum << endl;
}

```