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

#pragma GCC optimize ("Ofast")

#define all(x) x.begin() , x.end()
#define mp make_pair
#define gcd __gcd
typedef long long int ll;
typedef pair<ll , ll> pll;
typedef pair<int , int> pii;
typedef long double db;
typedef pair<ll , pll> plll;
typedef pair<int , pii> piii;
typedef pair<pll , pll> pllll;

const ll maxn = 1e3 + 16 , md = 998244353 , inf = 2e18;

ll tav(ll n , ll k){
	ll res = 1;
	while(k > 0){
		if(k & 1){
			res *= n; res %= md;
		}
		n *= n; n %= md;
		k >>= 1;
	}
	return res;
}

ll dp[maxn][maxn][2] , pd[maxn][maxn][2] , cd[maxn][maxn][2] , ad[maxn] , bd[maxn];

int main(){ // check MAXN
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	memset(dp , 0 , sizeof(dp));
	ll n , m;
	string a , b;
	cin>>a>>b; n = a.size(); m = b.size();
	ad[0] = 1;
	for(ll i = 1 ; i < n ; i++){
		ad[i] = ad[i - 1] * (a[i] != a[i - 1]) + 1;
	}
	bd[0] = 1;
	for(ll j = 1 ; j < m ; j++){
		bd[j] = bd[j - 1] * (b[j] != b[j - 1]) + 1;
	}	
	for(ll i = 0 ; i < n ; i++){
		pd[i][0][0] = pd[i][0][1] = (a[i] != b[0]);
		for(ll j = 1 ; j < m ; j++){
			pd[i][j][0] = bd[j] * (a[i] != b[j]);
			pd[i][j][1] = pd[i][j - 1][1] * (b[j] != b[j - 1]) + pd[i][j - 1][0] * (a[i] != b[j]) + (a[i] != b[j]);
		}
	}
	for(ll j = 0 ; j < m ; j++){
		cd[0][j][0] = cd[0][j][1] = (b[j] != a[0]);
		for(ll i = 1 ; i < n ; i++){
			cd[i][j][1] = ad[i] * (a[i] != b[j]);
			cd[i][j][0] = cd[i - 1][j][0] * (a[i] != a[i - 1]) + cd[i - 1][j][1] * (a[i] != b[j]) + (a[i] != b[j]);
		}
	}
	dp[0][0][0] = dp[0][0][1] = (a[0] != b[0]);
	for(ll j = 1 ; j < m ; j++){
		dp[0][j][1] = pd[0][j][1];
		dp[0][j][0] = pd[0][j][0];
	}
	for(ll i = 1 ; i < n ; i++){
		dp[i][0][0] = cd[i][0][0];
		dp[i][0][1] = cd[i][0][1];
		for(ll j = 1 ; j < m ; j++){
			dp[i][j][0] = dp[i - 1][j][0] * (a[i] != a[i - 1]) + dp[i - 1][j][1] * (a[i] != b[j]) + pd[i][j][0]; dp[i][j][0] %= md;
			dp[i][j][1] = dp[i][j - 1][0] * (a[i] != b[j]) + dp[i][j - 1][1] * (b[j] != b[j - 1]) + cd[i][j][1]; dp[i][j][1] %= md;
		}
	}
	ll ans = 0;
	for(ll i = 0 ; i < n ; i++){
		for(ll j = 0 ; j < m ; j++){
			ans += dp[i][j][0] + dp[i][j][1];
//			cout<<dp[i][j][0] + dp[i][j][1]<<' ';
		}
//		cout<<'\n';
	}
	ans %= md;
	cout<<ans<<'\n';
	return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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>
#include<fstream>
 
using namespace std;

#pragma GCC optimize ("Ofast")

#define all(x) x.begin() , x.end()
#define mp make_pair
#define gcd __gcd
typedef long long int ll;
typedef pair<ll , ll> pll;
typedef pair<int , int> pii;
typedef long double db;
typedef pair<ll , pll> plll;
typedef pair<int , pii> piii;
typedef pair<pll , pll> pllll;

const ll maxn = 1e3 + 16 , md = 998244353 , inf = 2e18;

ll tav(ll n , ll k){
	ll res = 1;
	while(k > 0){
		if(k & 1){
			res *= n; res %= md;
		}
		n *= n; n %= md;
		k >>= 1;
	}
	return res;
}

ll dp[maxn][maxn][2] , pd[maxn][maxn][2] , cd[maxn][maxn][2] , ad[maxn] , bd[maxn];

int main(){ // check MAXN
	ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	memset(dp , 0 , sizeof(dp));
	ll n , m;
	string a , b;
	cin>>a>>b; n = a.size(); m = b.size();
	ad[0] = 1;
	for(ll i = 1 ; i < n ; i++){
		ad[i] = ad[i - 1] * (a[i] != a[i - 1]) + 1;
	}
	bd[0] = 1;
	for(ll j = 1 ; j < m ; j++){
		bd[j] = bd[j - 1] * (b[j] != b[j - 1]) + 1;
	}	
	for(ll i = 0 ; i < n ; i++){
		pd[i][0][0] = pd[i][0][1] = (a[i] != b[0]);
		for(ll j = 1 ; j < m ; j++){
			pd[i][j][0] = bd[j] * (a[i] != b[j]);
			pd[i][j][1] = pd[i][j - 1][1] * (b[j] != b[j - 1]) + pd[i][j - 1][0] * (a[i] != b[j]) + (a[i] != b[j]);
		}
	}
	for(ll j = 0 ; j < m ; j++){
		cd[0][j][0] = cd[0][j][1] = (b[j] != a[0]);
		for(ll i = 1 ; i < n ; i++){
			cd[i][j][1] = ad[i] * (a[i] != b[j]);
			cd[i][j][0] = cd[i - 1][j][0] * (a[i] != a[i - 1]) + cd[i - 1][j][1] * (a[i] != b[j]) + (a[i] != b[j]);
		}
	}
	dp[0][0][0] = dp[0][0][1] = (a[0] != b[0]);
	for(ll j = 1 ; j < m ; j++){
		dp[0][j][1] = pd[0][j][1];
		dp[0][j][0] = pd[0][j][0];
	}
	for(ll i = 1 ; i < n ; i++){
		dp[i][0][0] = cd[i][0][0];
		dp[i][0][1] = cd[i][0][1];
		for(ll j = 1 ; j < m ; j++){
			dp[i][j][0] = dp[i - 1][j][0] * (a[i] != a[i - 1]) + dp[i - 1][j][1] * (a[i] != b[j]) + pd[i][j][0]; dp[i][j][0] %= md;
			dp[i][j][1] = dp[i][j - 1][0] * (a[i] != b[j]) + dp[i][j - 1][1] * (b[j] != b[j - 1]) + cd[i][j][1]; dp[i][j][1] %= md;
		}
	}
	ll ans = 0;
	for(ll i = 0 ; i < n ; i++){
		for(ll j = 0 ; j < m ; j++){
			ans += dp[i][j][0] + dp[i][j][1];
//			cout<<dp[i][j][0] + dp[i][j][1]<<' ';
		}
//		cout<<'\n';
	}
	ans %= md;
	cout<<ans<<'\n';
	return 0;
}
```