#define LOCAL
#include <bits/stdc++.h>
#include <assert.h>

using namespace std;

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define trav(a, x) for (auto& a: x)

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

using namespace std;

template <typename A, typename B>
ostream& operator <<(ostream& out, const pair<A, B>& a) {
  out << "(" << a.first << "," << a.second << ")";
  return out;
}
template <typename T, size_t N>
ostream& operator <<(ostream& out, const array<T, N>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T>
ostream& operator <<(ostream& out, const vector<T>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream& out, const set<T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
  return out;
}
template <typename U, typename T, class Cmp>
ostream& operator <<(ostream& out, const map<U, T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& p : a) { out << (first ? "" : ", "); out << p.first << ":" << p.second; first = 0;} out << "}";
  return out;
}

#ifdef LOCAL
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...) 42
#endif
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int mod = 998244353;
ll add(ll a, ll b){
	return (a + b)%mod;
}
void solve() {
	// once you know the tag or dp definition, you should be able to get answer
	// after that, try every possibilities
	int n, m; string x, y; cin >> x >> y;
	n = x.size();
	m = y.size();
	vector<vector<vector<ll>>> dp(n + 1, vector<vector<ll>>(m + 1, vector<ll>(2)));
	ll ans = 0;
	for (int i = 0; i <= n; i++){
		for (int j = 0; j <= m; j++){
			// take them as the first char;
			if (i < n) dp[i + 1][j][0] = add(dp[i + 1][j][0], 1);
			if (j < m) dp[i][j + 1][1] = add(dp[i][j + 1][1], 1);
			if (i > 0 && i < n && x[i] != x[i - 1]) dp[i + 1][j][0] = add(dp[i + 1][j][0], (dp[i][j][0]));
			if (i > 0 && j < m && y[j] != x[i - 1]) dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][0]);
			if (j > 0 && i < n && x[i] != y[j - 1]) dp[i + 1][j][0] = add(dp[i + 1][j][0],dp[i][j][1]);
			if (j > 0 && j < m && y[j] != y[j - 1]) dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][1]);
			ans = add(dp[i][j][0], ans);
			ans = add(dp[i][j][1], ans);
		}
	}
	int i = 0;
	for (; i < n; i++){
		int j = i;
		while (j + 1 < n && x[j + 1] != x[j]) j++;
		ll len = j + 1 - i;
		ll cnt = (len*(len + 1)/2)*(m + 1)%mod;
		ans = add(ans, -cnt);
		i = j;
	}
	int j = 0;
	for (; j < m; j++){
		int i = j;
		while (i + 1 < m && y[i + 1] != y[i]) i++;
		int len = i + 1 - j;
		ll cnt = (len*(len + 1)/2)*(n + 1)%mod;
		ans = add(ans, -cnt);
		j = i;
	}
	cout << ans << endl;
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int nofcase = 1;
	// cin >> nofcase;
	for (int tt = 1; tt <= nofcase; tt++) {
		// cout << "Case #" << tt << ": ";
		solve();
	}
	return 0;
}
### Prompt

Generate 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
#define LOCAL
#include <bits/stdc++.h>
#include <assert.h>

using namespace std;

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) x.begin(), x.end()
#define sz(x) (int)(x).size()
#define trav(a, x) for (auto& a: x)

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

using namespace std;

template <typename A, typename B>
ostream& operator <<(ostream& out, const pair<A, B>& a) {
  out << "(" << a.first << "," << a.second << ")";
  return out;
}
template <typename T, size_t N>
ostream& operator <<(ostream& out, const array<T, N>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T>
ostream& operator <<(ostream& out, const vector<T>& a) {
  out << "["; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "]";
  return out;
}
template <typename T, class Cmp>
ostream& operator <<(ostream& out, const set<T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& v : a) { out << (first ? "" : ", "); out << v; first = 0;} out << "}";
  return out;
}
template <typename U, typename T, class Cmp>
ostream& operator <<(ostream& out, const map<U, T, Cmp>& a) {
  out << "{"; bool first = true;
  for (auto& p : a) { out << (first ? "" : ", "); out << p.first << ":" << p.second; first = 0;} out << "}";
  return out;
}

#ifdef LOCAL
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...) 42
#endif
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
  cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
const int mod = 998244353;
ll add(ll a, ll b){
	return (a + b)%mod;
}
void solve() {
	// once you know the tag or dp definition, you should be able to get answer
	// after that, try every possibilities
	int n, m; string x, y; cin >> x >> y;
	n = x.size();
	m = y.size();
	vector<vector<vector<ll>>> dp(n + 1, vector<vector<ll>>(m + 1, vector<ll>(2)));
	ll ans = 0;
	for (int i = 0; i <= n; i++){
		for (int j = 0; j <= m; j++){
			// take them as the first char;
			if (i < n) dp[i + 1][j][0] = add(dp[i + 1][j][0], 1);
			if (j < m) dp[i][j + 1][1] = add(dp[i][j + 1][1], 1);
			if (i > 0 && i < n && x[i] != x[i - 1]) dp[i + 1][j][0] = add(dp[i + 1][j][0], (dp[i][j][0]));
			if (i > 0 && j < m && y[j] != x[i - 1]) dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][0]);
			if (j > 0 && i < n && x[i] != y[j - 1]) dp[i + 1][j][0] = add(dp[i + 1][j][0],dp[i][j][1]);
			if (j > 0 && j < m && y[j] != y[j - 1]) dp[i][j + 1][1] = add(dp[i][j + 1][1], dp[i][j][1]);
			ans = add(dp[i][j][0], ans);
			ans = add(dp[i][j][1], ans);
		}
	}
	int i = 0;
	for (; i < n; i++){
		int j = i;
		while (j + 1 < n && x[j + 1] != x[j]) j++;
		ll len = j + 1 - i;
		ll cnt = (len*(len + 1)/2)*(m + 1)%mod;
		ans = add(ans, -cnt);
		i = j;
	}
	int j = 0;
	for (; j < m; j++){
		int i = j;
		while (i + 1 < m && y[i + 1] != y[i]) i++;
		int len = i + 1 - j;
		ll cnt = (len*(len + 1)/2)*(n + 1)%mod;
		ans = add(ans, -cnt);
		j = i;
	}
	cout << ans << endl;
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int nofcase = 1;
	// cin >> nofcase;
	for (int tt = 1; tt <= nofcase; tt++) {
		// cout << "Case #" << tt << ": ";
		solve();
	}
	return 0;
}
```