#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define trace(...) debug(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void debug(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void debug(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " | ";
  debug(comma + 1, args...);
}

const int INF=1000000000+5;
const int N=1000+5;
const int M=5e5+5;
const ll oo=1e18+5;
const int mod=998244353;

int dp[N][N][3][4];
string x,y;

void add_self(int& a,int b){
	a+=b;
	if(a>=mod) a-=mod;
}

int calc(int a,int b,int last,int mask){
	//~ trace(dp[a][b][last][mask]);
	if(dp[a][b][last][mask]!=-1) return dp[a][b][last][mask];
	char prev;
	int ans=0;
	if(mask==3) ans=1;
	if(last==2){
		prev='z'+1;
	}
	else{
		if(!last) prev=x[a-1];
		else prev=y[b-1];
	}
	if(a<(int)x.size()&&x[a]!=prev){
		add_self(ans,calc(a+1,b,0,mask|1));
	}
	if(b<(int)y.size()&&y[b]!=prev){
		add_self(ans,calc(a,b+1,1,mask|2));
	}
	//~ trace(ans);
	return dp[a][b][last][mask]=ans;
}

int main(){
	ios::sync_with_stdio(0); cin.tie(0);
	cin>>x>>y;
	int ans=0;
	//~ for(int i=0;i<N;++i){
		//~ for(int j=0;j<N;++j){
			//~ for(int last=0;last<2;++last){
				//~ for(int mask=0;mask<4;++mask) dp[i][j][last][mask]=-1;
			//~ }
		//~ }
	//~ }
	memset(dp,-1,sizeof dp);
	for(int i=0;i<(int)x.size();++i){
		for(int j=0;j<(int)y.size();++j){
			add_self(ans,calc(i,j,2,0));
			//~ trace(calc(i,j,2,0));
		}
	}
	cout<<ans;
	return 0;
}

### Prompt

In Cpp, your task is to solve 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;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
#define trace(...) debug(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void debug(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void debug(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " | ";
  debug(comma + 1, args...);
}

const int INF=1000000000+5;
const int N=1000+5;
const int M=5e5+5;
const ll oo=1e18+5;
const int mod=998244353;

int dp[N][N][3][4];
string x,y;

void add_self(int& a,int b){
	a+=b;
	if(a>=mod) a-=mod;
}

int calc(int a,int b,int last,int mask){
	//~ trace(dp[a][b][last][mask]);
	if(dp[a][b][last][mask]!=-1) return dp[a][b][last][mask];
	char prev;
	int ans=0;
	if(mask==3) ans=1;
	if(last==2){
		prev='z'+1;
	}
	else{
		if(!last) prev=x[a-1];
		else prev=y[b-1];
	}
	if(a<(int)x.size()&&x[a]!=prev){
		add_self(ans,calc(a+1,b,0,mask|1));
	}
	if(b<(int)y.size()&&y[b]!=prev){
		add_self(ans,calc(a,b+1,1,mask|2));
	}
	//~ trace(ans);
	return dp[a][b][last][mask]=ans;
}

int main(){
	ios::sync_with_stdio(0); cin.tie(0);
	cin>>x>>y;
	int ans=0;
	//~ for(int i=0;i<N;++i){
		//~ for(int j=0;j<N;++j){
			//~ for(int last=0;last<2;++last){
				//~ for(int mask=0;mask<4;++mask) dp[i][j][last][mask]=-1;
			//~ }
		//~ }
	//~ }
	memset(dp,-1,sizeof dp);
	for(int i=0;i<(int)x.size();++i){
		for(int j=0;j<(int)y.size();++j){
			add_self(ans,calc(i,j,2,0));
			//~ trace(calc(i,j,2,0));
		}
	}
	cout<<ans;
	return 0;
}

```