#include<bits/stdc++.h>
#define S second 
#define F first
#define all(x) x.begin(),x.end()
#define debug(x) cout<<x<<":"<<#x<<"\n";
using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
typedef long double ld;
ll inf = 1000000ll*1000000*1000000*4;
ll mod = 998244353;
const int maxn = 1024;
int Ls[maxn];
int Lt[maxn];
string s,t;
ll dp[maxn][maxn][2],ans;
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>s>>t;
	int last = 0;
	Ls[0] = 1;
	Lt[0] = 1;
	for(int i=1;i<s.size();i++){
		if(s[i] == s[i-1])last = i;
		Ls[i] = i-last+1;
	}
	last = 0;
	for(int i=1;i<t.size();i++){
		if(t[i] == t[i-1])last = i;
		Lt[i] = i-last+1;
	}
	/*for(int i=0;i<s.size();i++){
		cout<<Ls[i]<<" ";
	}cout<<"\n";
	for(int i=0;i<t.size();i++){
		cout<<Lt[i]<<" ";
	}cout<<"\n";*/
	for(int i=0;i<s.size();i++){
		for(int j=0;j<t.size();j++){
			if(i){
				dp[i][j][0] += dp[i-1][j][0]*(s[i] != s[i-1])+dp[i-1][j][1]*(s[i] != t[j]);
			}
			if(j){
				dp[i][j][1] += dp[i][j-1][1]*(t[j] != t[j-1])+dp[i][j-1][0]*(s[i] != t[j]);
			}
			dp[i][j][0] += Lt[j]*(s[i] != t[j]);
			dp[i][j][1] += Ls[i]*(s[i] != t[j]);
			dp[i][j][0] %= mod;
			dp[i][j][1] %= mod;
			ans += dp[i][j][0]+dp[i][j][1];
			ans %= mod;
		}
	}
	cout<<ans;
	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>
#define S second 
#define F first
#define all(x) x.begin(),x.end()
#define debug(x) cout<<x<<":"<<#x<<"\n";
using namespace std;
typedef long long int ll;
typedef pair<int,int> pii;
typedef long double ld;
ll inf = 1000000ll*1000000*1000000*4;
ll mod = 998244353;
const int maxn = 1024;
int Ls[maxn];
int Lt[maxn];
string s,t;
ll dp[maxn][maxn][2],ans;
int main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>s>>t;
	int last = 0;
	Ls[0] = 1;
	Lt[0] = 1;
	for(int i=1;i<s.size();i++){
		if(s[i] == s[i-1])last = i;
		Ls[i] = i-last+1;
	}
	last = 0;
	for(int i=1;i<t.size();i++){
		if(t[i] == t[i-1])last = i;
		Lt[i] = i-last+1;
	}
	/*for(int i=0;i<s.size();i++){
		cout<<Ls[i]<<" ";
	}cout<<"\n";
	for(int i=0;i<t.size();i++){
		cout<<Lt[i]<<" ";
	}cout<<"\n";*/
	for(int i=0;i<s.size();i++){
		for(int j=0;j<t.size();j++){
			if(i){
				dp[i][j][0] += dp[i-1][j][0]*(s[i] != s[i-1])+dp[i-1][j][1]*(s[i] != t[j]);
			}
			if(j){
				dp[i][j][1] += dp[i][j-1][1]*(t[j] != t[j-1])+dp[i][j-1][0]*(s[i] != t[j]);
			}
			dp[i][j][0] += Lt[j]*(s[i] != t[j]);
			dp[i][j][1] += Ls[i]*(s[i] != t[j]);
			dp[i][j][0] %= mod;
			dp[i][j][1] %= mod;
			ans += dp[i][j][0]+dp[i][j][1];
			ans %= mod;
		}
	}
	cout<<ans;
	return 0;
}

```