#include "bits/stdc++.h"
using namespace std;
#define pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll

template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.fr>>a.sc;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.fr<<" "<<a.sc;return out;}
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}

const long long INF=1e18;
const int32_t M=998244353;

const int N=1005;

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

int calc(int a,int b,int last,int started){
	if(dp[a][b][last][started] != -1){
		return dp[a][b][last][started];
	}
	int ans = (started == 3);
	char pc;
	if(last == 2){
		pc = 'z' + 1;
	}
	else{
		pc = ((last == 0)?x[a-1]:y[b-1]);
	}
	
	if(a < sz(x) && x[a] != pc){
		ans = (ans + calc(a+1,b,0,started | 1)) % M;
	}
	
	if(b < sz(y) && y[b] != pc){
		ans = (ans + calc(a,b+1,1,started | 2)) % M;
	}
	return dp[a][b][last][started] = ans;
}

void solve(){
	mem1(dp);
	cin >> x >> y;
	
	int ans = 0;
	rep(i,0,sz(x)){
		rep(j,0,sz(y)){
			
			ans = (ans + calc(i,j,2,0)) % M;
		}
	}
	cout << ans << "\n";
}
signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	#ifdef SIEVE
		sieve();
	#endif
	#ifdef NCR
		init();
	#endif
	int t=1;
	//cin>>t;
	while(t--) solve();
	return 0;
}

### Prompt

Please create a solution in Cpp 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 pb                push_back
#define ppb               pop_back
#define pf                push_front
#define ppf               pop_front
#define all(x)            (x).begin(),(x).end()
#define uniq(v)           (v).erase(unique(all(v)),(v).end())
#define sz(x)             (int)((x).size())
#define fr                first
#define sc                second
#define pii               pair<int,int>
#define rep(i,a,b)        for(int i=a;i<b;i++)
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))
#define ppc               __builtin_popcount
#define ppcll             __builtin_popcountll

template<typename T1,typename T2>istream& operator>>(istream& in,pair<T1,T2> &a){in>>a.fr>>a.sc;return in;}
template<typename T1,typename T2>ostream& operator<<(ostream& out,pair<T1,T2> a){out<<a.fr<<" "<<a.sc;return out;}
template<typename T,typename T1>T amax(T &a,T1 b){if(b>a)a=b;return a;}
template<typename T,typename T1>T amin(T &a,T1 b){if(b<a)a=b;return a;}

const long long INF=1e18;
const int32_t M=998244353;

const int N=1005;

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

int calc(int a,int b,int last,int started){
	if(dp[a][b][last][started] != -1){
		return dp[a][b][last][started];
	}
	int ans = (started == 3);
	char pc;
	if(last == 2){
		pc = 'z' + 1;
	}
	else{
		pc = ((last == 0)?x[a-1]:y[b-1]);
	}
	
	if(a < sz(x) && x[a] != pc){
		ans = (ans + calc(a+1,b,0,started | 1)) % M;
	}
	
	if(b < sz(y) && y[b] != pc){
		ans = (ans + calc(a,b+1,1,started | 2)) % M;
	}
	return dp[a][b][last][started] = ans;
}

void solve(){
	mem1(dp);
	cin >> x >> y;
	
	int ans = 0;
	rep(i,0,sz(x)){
		rep(j,0,sz(y)){
			
			ans = (ans + calc(i,j,2,0)) % M;
		}
	}
	cout << ans << "\n";
}
signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);cout.tie(0);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	#ifdef SIEVE
		sieve();
	#endif
	#ifdef NCR
		init();
	#endif
	int t=1;
	//cin>>t;
	while(t--) solve();
	return 0;
}

```