#include<bits/stdc++.h>
using namespace std;
#define maxn 1013
#define M 998244353
#define ll long long

ll dp[maxn][maxn][2][2][2];
ll res;
string a, b;
int main(){
	cin>>a>>b;
	int n1=(int)a.size(), n2=(int)b.size();
	a="*"+a, b="*"+b;
	for(int i=0; i<=n1; i++){
		for(int j=0; j<=n2; j++){
			if(i && a[i]!=a[i-1]){
				dp[i][j][0][1][0]+=dp[i-1][j][0][1][0];
				dp[i][j][0][1][1]+=dp[i-1][j][0][1][1];
			}
			if(j && b[j]!=b[j-1]){
				dp[i][j][1][0][1]+=dp[i][j-1][1][0][1];
				dp[i][j][1][1][1]+=dp[i][j-1][1][1][1];
			}
			if(i && j && a[i]!=b[j]){
				dp[i][j][0][1][1]+=dp[i-1][j][1][0][1]+dp[i-1][j][1][1][1];
				dp[i][j][1][1][1]+=dp[i][j-1][0][1][0]+dp[i][j-1][0][1][1];
			}
			if(i) dp[i][j][0][1][0]++;
			if(j) dp[i][j][1][0][1]++;
			for(int k1=0; k1<2; k1++)
				for(int k2=0; k2<2; k2++)
					for(int k3=0; k3<2; k3++)
						dp[i][j][k1][k2][k3]%=M;
		//	cout<<dp[i][j][0][1][1]<<" "<<dp[i][j][1][1][1]<<"\n";
			res+=dp[i][j][0][1][1]+dp[i][j][1][1][1];
			res%=M;
		}
	}
	cout<<res;
	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>
using namespace std;
#define maxn 1013
#define M 998244353
#define ll long long

ll dp[maxn][maxn][2][2][2];
ll res;
string a, b;
int main(){
	cin>>a>>b;
	int n1=(int)a.size(), n2=(int)b.size();
	a="*"+a, b="*"+b;
	for(int i=0; i<=n1; i++){
		for(int j=0; j<=n2; j++){
			if(i && a[i]!=a[i-1]){
				dp[i][j][0][1][0]+=dp[i-1][j][0][1][0];
				dp[i][j][0][1][1]+=dp[i-1][j][0][1][1];
			}
			if(j && b[j]!=b[j-1]){
				dp[i][j][1][0][1]+=dp[i][j-1][1][0][1];
				dp[i][j][1][1][1]+=dp[i][j-1][1][1][1];
			}
			if(i && j && a[i]!=b[j]){
				dp[i][j][0][1][1]+=dp[i-1][j][1][0][1]+dp[i-1][j][1][1][1];
				dp[i][j][1][1][1]+=dp[i][j-1][0][1][0]+dp[i][j-1][0][1][1];
			}
			if(i) dp[i][j][0][1][0]++;
			if(j) dp[i][j][1][0][1]++;
			for(int k1=0; k1<2; k1++)
				for(int k2=0; k2<2; k2++)
					for(int k3=0; k3<2; k3++)
						dp[i][j][k1][k2][k3]%=M;
		//	cout<<dp[i][j][0][1][1]<<" "<<dp[i][j][1][1][1]<<"\n";
			res+=dp[i][j][0][1][1]+dp[i][j][1][1][1];
			res%=M;
		}
	}
	cout<<res;
	return 0;
}

```