#include<bits/stdc++.h>
using namespace std;

const int mod=998244353;

const int maxn=1005;
int m,n,ans;
string x,y;
int f[maxn][maxn][2][4];

void add(int &t1,int t2) {
	t1%=mod; t2%=mod;
	t1+=t2;
	t1%=mod;
}

int main() {
	cin>>x>>y;
	n=x.size(); m=y.size();
	x=' '+x; y=' '+y;
	for(int i=0;i<=n;i++)
		for(int j=0;j<=m;j++) {
			if(i<n) add(f[i+1][j][0][1],1);
			if(j<m) add(f[i][j+1][1][2],1);
			for(int msk=0;msk<4;msk++) {
				if(i<n&&x[i]!=x[i+1]) {
					add(f[i+1][j][0][msk|1],f[i][j][0][msk]);
				}
				if(j<m&&x[i]!=y[j+1]) {
					add(f[i][j+1][1][msk|2],f[i][j][0][msk]);
				}
				if(j<m&&y[j]!=y[j+1]) {
					add(f[i][j+1][1][msk|2],f[i][j][1][msk]);
				}
				if(i<n&&x[i+1]!=y[j]) {
					add(f[i+1][j][0][msk|1],f[i][j][1][msk]);
				}
			}
		}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++) {
			add(ans,f[i][j][0][3]);
			add(ans,f[i][j][1][3]);
		}
	cout<<ans<<endl;
	return 0;
}
### Prompt

Create a solution in cpp for 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;

const int mod=998244353;

const int maxn=1005;
int m,n,ans;
string x,y;
int f[maxn][maxn][2][4];

void add(int &t1,int t2) {
	t1%=mod; t2%=mod;
	t1+=t2;
	t1%=mod;
}

int main() {
	cin>>x>>y;
	n=x.size(); m=y.size();
	x=' '+x; y=' '+y;
	for(int i=0;i<=n;i++)
		for(int j=0;j<=m;j++) {
			if(i<n) add(f[i+1][j][0][1],1);
			if(j<m) add(f[i][j+1][1][2],1);
			for(int msk=0;msk<4;msk++) {
				if(i<n&&x[i]!=x[i+1]) {
					add(f[i+1][j][0][msk|1],f[i][j][0][msk]);
				}
				if(j<m&&x[i]!=y[j+1]) {
					add(f[i][j+1][1][msk|2],f[i][j][0][msk]);
				}
				if(j<m&&y[j]!=y[j+1]) {
					add(f[i][j+1][1][msk|2],f[i][j][1][msk]);
				}
				if(i<n&&x[i+1]!=y[j]) {
					add(f[i+1][j][0][msk|1],f[i][j][1][msk]);
				}
			}
		}
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++) {
			add(ans,f[i][j][0][3]);
			add(ans,f[i][j][1][3]);
		}
	cout<<ans<<endl;
	return 0;
}
```