#include<bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define all(v) v.begin(),v.end()
#define sz(x) (ll)x.size()
#define F first
#define S second
typedef long long ll;
using namespace std;

const int N = 1001;
const int mod = 998244353;

int n, m;
int x, y;
string s1, s2;
int dp[2][2][3][N][N];

int solve(bool use1, bool use2, int prev, int l1, int l2) {
	if(!use1 && !use2)
		return 1;
	int& ret = dp[use1][use2][prev][l1][l2];
	if(ret != -1)
		return ret;
	char prevc = '`';
	switch (prev) {
		case 1:
			prevc = s1[l1 - 1];
			break;
		case 2:
			prevc = s2[l2 - 1];
			break;
	}
	ret = 0;
	if(use1 && l1 < n && s1[l1] != prevc) {
		ret = (ret + solve(true, use2, 1, l1 + 1, l2)) % mod;
		ret = (ret + solve(false, use2, 1, l1 + 1, l2)) % mod;
	}
	if(use2 && l2 < m && s2[l2] != prevc) {
		ret = (ret + solve(use1, true, 2, l1, l2 + 1)) % mod;
		ret = (ret + solve(use1, false, 2, l1, l2 + 1)) % mod;
	}
	return ret;
}

int main(){
    IOS

    memset(dp, -1, sizeof(dp));
    cin >> s1 >> s2;
    n = s1.length();
    m = s2.length();
    int ans = 0;
    for(int i = 0; i < n; ++i) {
    	for(int j = 0; j < m; ++j) {
    		x = i, y = j;
    		ans = (ans + solve(true, true, 0, i, j)) % mod;
    	}
    }
    cout << ans;

    return 0;
}
### Prompt

Please formulate a Cpp solution to the following problem:

You are given two strings x and y, both consist only of lowercase Latin letters. Let |s| be the length of string s.

Let's call a sequence a a merging sequence if it consists of exactly |x| zeros and exactly |y| ones in some order.

A merge z is produced from a sequence a by the following rules: 

  * if a_i=0, then remove a letter from the beginning of x and append it to the end of z; 
  * if a_i=1, then remove a letter from the beginning of y and append it to the end of z. 



Two merging sequences a and b are different if there is some position i such that a_i ≠ b_i.

Let's call a string z chaotic if for all i from 2 to |z| z_{i-1} ≠ z_i.

Let s[l,r] for some 1 ≤ l ≤ r ≤ |s| be a substring of consecutive letters of s, starting from position l and ending at position r inclusive.

Let f(l_1, r_1, l_2, r_2) be the number of different merging sequences of x[l_1,r_1] and y[l_2,r_2] that produce chaotic merges. Note that only non-empty substrings of x and y are considered.

Calculate ∑ _{1 ≤ l_1 ≤ r_1 ≤ |x| \\\ 1 ≤ l_2 ≤ r_2 ≤ |y|} f(l_1, r_1, l_2, r_2). Output the answer modulo 998 244 353.

Input

The first line contains a string x (1 ≤ |x| ≤ 1000).

The second line contains a string y (1 ≤ |y| ≤ 1000).

Both strings consist only of lowercase Latin letters.

Output

Print a single integer — the sum of f(l_1, r_1, l_2, r_2) over 1 ≤ l_1 ≤ r_1 ≤ |x| and 1 ≤ l_2 ≤ r_2 ≤ |y| modulo 998 244 353.

Examples

Input


aaa
bb


Output


24


Input


code
forces


Output


1574


Input


aaaaa
aaa


Output


0


Input


justamassivetesttocheck
howwellyouhandlemodulooperations


Output


667387032

Note

In the first example there are: 

  * 6 pairs of substrings "a" and "b", each with valid merging sequences "01" and "10"; 
  * 3 pairs of substrings "a" and "bb", each with a valid merging sequence "101"; 
  * 4 pairs of substrings "aa" and "b", each with a valid merging sequence "010"; 
  * 2 pairs of substrings "aa" and "bb", each with valid merging sequences "0101" and "1010"; 
  * 2 pairs of substrings "aaa" and "b", each with no valid merging sequences; 
  * 1 pair of substrings "aaa" and "bb" with a valid merging sequence "01010"; 



Thus, the answer is 6 ⋅ 2 + 3 ⋅ 1 + 4 ⋅ 1 + 2 ⋅ 2 + 2 ⋅ 0 + 1 ⋅ 1 = 24.

### Response

```cpp
#include<bits/stdc++.h>
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define all(v) v.begin(),v.end()
#define sz(x) (ll)x.size()
#define F first
#define S second
typedef long long ll;
using namespace std;

const int N = 1001;
const int mod = 998244353;

int n, m;
int x, y;
string s1, s2;
int dp[2][2][3][N][N];

int solve(bool use1, bool use2, int prev, int l1, int l2) {
	if(!use1 && !use2)
		return 1;
	int& ret = dp[use1][use2][prev][l1][l2];
	if(ret != -1)
		return ret;
	char prevc = '`';
	switch (prev) {
		case 1:
			prevc = s1[l1 - 1];
			break;
		case 2:
			prevc = s2[l2 - 1];
			break;
	}
	ret = 0;
	if(use1 && l1 < n && s1[l1] != prevc) {
		ret = (ret + solve(true, use2, 1, l1 + 1, l2)) % mod;
		ret = (ret + solve(false, use2, 1, l1 + 1, l2)) % mod;
	}
	if(use2 && l2 < m && s2[l2] != prevc) {
		ret = (ret + solve(use1, true, 2, l1, l2 + 1)) % mod;
		ret = (ret + solve(use1, false, 2, l1, l2 + 1)) % mod;
	}
	return ret;
}

int main(){
    IOS

    memset(dp, -1, sizeof(dp));
    cin >> s1 >> s2;
    n = s1.length();
    m = s2.length();
    int ans = 0;
    for(int i = 0; i < n; ++i) {
    	for(int j = 0; j < m; ++j) {
    		x = i, y = j;
    		ans = (ans + solve(true, true, 0, i, j)) % mod;
    	}
    }
    cout << ans;

    return 0;
}
```