/*
Hey  why peeping here -_'_- ?
I believe on myself and I will achieve
this->author = Fuad Ashraful Mehmet, CSE ,University of Asia Pacific
Todo:
https://www.youtube.com/watch?v=VoTUc-1qIT8&ab_channel=TipsJhankarGaane
https://codeforces.com/problemset/page/5?tags=1500-2500
*/
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < (int)n; ++i)
#define all(o) (o).begin(), (o).end()
#define rall(o) (o).rbegin(), (o).rend()
#define sz(o) (o).size()
#define FAST_IO  ios_base::sync_with_stdio(0); cin.tie(nullptr); cout.tie(nullptr);
#define dbg(o) cerr << "at " << __LINE__ << " response = " << o << endl;
#define dbgA(a,L,R) cerr<<"A=[";for(int i=L;i<=R;i++) cerr<<a[i]<<(i==R?"]\n":" ");
#define dbgV(O) cerr << #O << "=["; for (auto o : O) cerr << o << ", "; cerr << "E]\n";
#define pb push_back
using namespace std;
using pii = pair<int, int>;
typedef long long ll;
int  n,m,k,tc=1,q,cs=1,lx,ly;
const int N=1005;

ll dp[N][N][3][2][2];
string x,y;
const ll mod=998244353;

void relax(ll &num)
{
	num%=mod;
	if(num<0)
	num+=mod;
	return;
}

ll go(int i,int j,int prev,bool f1,bool f2)
{
	//if(i>=lx or j>=ly) return 0;
	ll &ret=dp[i][j][prev][f1][f2];
	if(-1!=ret)
		return ret;
	
	if(prev==0)
		prev=x[i-1]-'a';
	else if(prev==1)
		prev=y[j-1]-'a';
	else prev=27;
	ret=f1&f2;
	if(i<lx)
	if(prev!=x[i]-'a')ret+=go(i+1,j,0,f1|1,f2);
	relax(ret);
	if(j<ly)
	if(prev!=y[j]-'a')ret+=go(i,j+1,1,f1,f2|1);
	relax(ret);

	return ret;
}
void Solve()
{
	memset(dp,-1,sizeof dp);
	cin>>x>>y;
	lx=sz(x);
	ly=sz(y);

	ll ans=0;
	forn(i,lx)
	{
		forn(j,ly)
		{
			ans+=go(i,j,2,0,0);
			relax(ans);
		}
	}

	cout<<ans<<'\n';
	return;
	
}

int main()
{
	FAST_IO
	//cout<<setprecision(20)<<fixed;
	//cin>>tc;
	while(tc--)
	{
		Solve();
	}
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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
/*
Hey  why peeping here -_'_- ?
I believe on myself and I will achieve
this->author = Fuad Ashraful Mehmet, CSE ,University of Asia Pacific
Todo:
https://www.youtube.com/watch?v=VoTUc-1qIT8&ab_channel=TipsJhankarGaane
https://codeforces.com/problemset/page/5?tags=1500-2500
*/
#include <bits/stdc++.h>
#define forn(i, n) for (int i = 0; i < (int)n; ++i)
#define all(o) (o).begin(), (o).end()
#define rall(o) (o).rbegin(), (o).rend()
#define sz(o) (o).size()
#define FAST_IO  ios_base::sync_with_stdio(0); cin.tie(nullptr); cout.tie(nullptr);
#define dbg(o) cerr << "at " << __LINE__ << " response = " << o << endl;
#define dbgA(a,L,R) cerr<<"A=[";for(int i=L;i<=R;i++) cerr<<a[i]<<(i==R?"]\n":" ");
#define dbgV(O) cerr << #O << "=["; for (auto o : O) cerr << o << ", "; cerr << "E]\n";
#define pb push_back
using namespace std;
using pii = pair<int, int>;
typedef long long ll;
int  n,m,k,tc=1,q,cs=1,lx,ly;
const int N=1005;

ll dp[N][N][3][2][2];
string x,y;
const ll mod=998244353;

void relax(ll &num)
{
	num%=mod;
	if(num<0)
	num+=mod;
	return;
}

ll go(int i,int j,int prev,bool f1,bool f2)
{
	//if(i>=lx or j>=ly) return 0;
	ll &ret=dp[i][j][prev][f1][f2];
	if(-1!=ret)
		return ret;
	
	if(prev==0)
		prev=x[i-1]-'a';
	else if(prev==1)
		prev=y[j-1]-'a';
	else prev=27;
	ret=f1&f2;
	if(i<lx)
	if(prev!=x[i]-'a')ret+=go(i+1,j,0,f1|1,f2);
	relax(ret);
	if(j<ly)
	if(prev!=y[j]-'a')ret+=go(i,j+1,1,f1,f2|1);
	relax(ret);

	return ret;
}
void Solve()
{
	memset(dp,-1,sizeof dp);
	cin>>x>>y;
	lx=sz(x);
	ly=sz(y);

	ll ans=0;
	forn(i,lx)
	{
		forn(j,ly)
		{
			ans+=go(i,j,2,0,0);
			relax(ans);
		}
	}

	cout<<ans<<'\n';
	return;
	
}

int main()
{
	FAST_IO
	//cout<<setprecision(20)<<fixed;
	//cin>>tc;
	while(tc--)
	{
		Solve();
	}
	return 0;
}
```