#include <bits/stdc++.h>
#pragma GCC optimize ("O3", "unroll-all-loops")
#pragma GCC target ("sse4.2")

using namespace std;

#define F first
#define S second
typedef long long       ll;
typedef long double     ld;
typedef pair<ll, ll>   pll; 
typedef pair<int, int> pii; 

ifstream in;
ofstream out;

const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = 998244353;
const long long inf = ml * ml * ml + 7; 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m;
bool viv = false;

vector<ll> get(string s) {
	int n = s.size();
	vector<ll> res(n + 1, 1);
	bool good = true;
	for (int i = 1; i < n; i++) {
		good &= (s[i] != s[i - 1]);
		res[i + 1] = good;
	}
	return res;
}

vector<ll> get_nice(string s) {
	int n = s.size();
	vector<ll> res(n, 1);
	for (int i = 1; i < n; i++)
		if (s[i] != s[i - 1])
			res[i] += res[i - 1];
	return res;
}

void solve() {
	string s, t;
	cin >> s >> t;
	n = s.size(), m = t.size();


	vector<ll> prefs = get(s);
	vector<ll> preft = get(t);
	vector<ll> nices = get_nice(s);
	vector<ll> nicet = get_nice(t);

	vector<vector<ll>> dp[2];
	dp[0].resize(n + 1, vector<ll>(m + 1, 0));
	dp[1].resize(m + 1, vector<ll>(n + 1, 0));

	for (int i = 0; i < n; i++)
		dp[0][i + 1][0] = nices[i];
	for (int i = 0; i < m; i++)
		dp[1][i + 1][0] = nicet[i];

	ll ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (i && s[i] != s[i - 1])
				dp[0][i + 1][j + 1] += dp[0][i][j + 1];
			if (s[i] != t[j])
				dp[0][i + 1][j + 1] += dp[1][j + 1][i];
			if (i && s[i] != t[j])
				dp[0][i + 1][j + 1] += nicet[j];
			dp[0][i + 1][j + 1] %= mod;
	
			if (j && t[j] != t[j - 1])
				dp[1][j + 1][i + 1] += dp[1][j][i + 1];
			if (t[j] != s[i])
				dp[1][j + 1][i + 1] += dp[0][i + 1][j];
			if (j && t[j] != s[i])
				dp[1][j + 1][i + 1] += nices[i];
			dp[1][j + 1][i + 1] %= mod;

			if (viv) {
	 			cout << "0 " << i + 1 << ' ' << j + 1 << ": " << dp[0][i + 1][j + 1] << endl;
				cout << "1 " << j + 1 << ' ' << i + 1 << ": " << dp[1][j + 1][i + 1] << endl;
			}

			ans += dp[0][i + 1][j + 1];
			ans += dp[1][j + 1][i + 1];	
			ans %= mod;
		}
	}
	cout << ans << '\n';
}

int main() {
	// viv = true;
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll t = 1; 
	// cin >> t;
	while (t--)
		solve();

	return 0;
}
### Prompt

Your task is to create a CPP solution 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>
#pragma GCC optimize ("O3", "unroll-all-loops")
#pragma GCC target ("sse4.2")

using namespace std;

#define F first
#define S second
typedef long long       ll;
typedef long double     ld;
typedef pair<ll, ll>   pll; 
typedef pair<int, int> pii; 

ifstream in;
ofstream out;

const long long kk = 1000;
const long long ml = kk * kk;
const long long mod = 998244353;
const long long inf = ml * ml * ml + 7; 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m;
bool viv = false;

vector<ll> get(string s) {
	int n = s.size();
	vector<ll> res(n + 1, 1);
	bool good = true;
	for (int i = 1; i < n; i++) {
		good &= (s[i] != s[i - 1]);
		res[i + 1] = good;
	}
	return res;
}

vector<ll> get_nice(string s) {
	int n = s.size();
	vector<ll> res(n, 1);
	for (int i = 1; i < n; i++)
		if (s[i] != s[i - 1])
			res[i] += res[i - 1];
	return res;
}

void solve() {
	string s, t;
	cin >> s >> t;
	n = s.size(), m = t.size();


	vector<ll> prefs = get(s);
	vector<ll> preft = get(t);
	vector<ll> nices = get_nice(s);
	vector<ll> nicet = get_nice(t);

	vector<vector<ll>> dp[2];
	dp[0].resize(n + 1, vector<ll>(m + 1, 0));
	dp[1].resize(m + 1, vector<ll>(n + 1, 0));

	for (int i = 0; i < n; i++)
		dp[0][i + 1][0] = nices[i];
	for (int i = 0; i < m; i++)
		dp[1][i + 1][0] = nicet[i];

	ll ans = 0;
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			if (i && s[i] != s[i - 1])
				dp[0][i + 1][j + 1] += dp[0][i][j + 1];
			if (s[i] != t[j])
				dp[0][i + 1][j + 1] += dp[1][j + 1][i];
			if (i && s[i] != t[j])
				dp[0][i + 1][j + 1] += nicet[j];
			dp[0][i + 1][j + 1] %= mod;
	
			if (j && t[j] != t[j - 1])
				dp[1][j + 1][i + 1] += dp[1][j][i + 1];
			if (t[j] != s[i])
				dp[1][j + 1][i + 1] += dp[0][i + 1][j];
			if (j && t[j] != s[i])
				dp[1][j + 1][i + 1] += nices[i];
			dp[1][j + 1][i + 1] %= mod;

			if (viv) {
	 			cout << "0 " << i + 1 << ' ' << j + 1 << ": " << dp[0][i + 1][j + 1] << endl;
				cout << "1 " << j + 1 << ' ' << i + 1 << ": " << dp[1][j + 1][i + 1] << endl;
			}

			ans += dp[0][i + 1][j + 1];
			ans += dp[1][j + 1][i + 1];	
			ans %= mod;
		}
	}
	cout << ans << '\n';
}

int main() {
	// viv = true;
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	ll t = 1; 
	// cin >> t;
	while (t--)
		solve();

	return 0;
}
```