#include <time.h>
#include <iomanip>
#include <iostream>
#include <map>
#include <algorithm>
#include <queue>
#include <set>
#include <vector>
#include <iomanip>
#include <bitset>
using namespace std;
long long kq = 0 , f[1005][1005][3] , mod = 998244353;
string x , y;
void trau(int sti , int ste , int i , int e , string s)
{
	if (s.size() > 1)
	{
		if (s[s.size() - 1] == s[s.size() - 2])
		{
			return;
		}
		if (i != sti && e != ste)
		{
			cout <<sti<<" "<<ste<<" "<<i<<" "<<e<<" "<<s<<'\n';
			kq++;
		}
	}
	if (i+1 < x.size())
	{
		s += x[i+1];
		trau(sti , ste , i + 1 , e , s);
		s.pop_back();
	}
	if (e+1 < y.size())
	{
		s += y[e+1];
		trau(sti , ste , i , e + 1 , s);
		s.pop_back();
	}
}
int tt()
{
	for (int i = 1 ; i < x.size() ; i++)
	{
		for (int e = 1 ; e < y.size() ; e++)
		{
			trau(i-1 , e-1 , i-1 , e-1 , "");
		}
	}
	return kq;
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	srand (time(NULL));
	int test = 1;
	cin >> x;
	cin >> y;
	int n = x.size();
	int m = y.size();
	x = '.' + x;
	y = '/' + y;
	//cout <<tt()<<'\n';
	f[1][0][0] = 1;
	f[0][1][1] = 1;
	long long res = 0;
	for (int i = 2 ; i <= n ; i++)
	{
		if (x[i] != x[i-1])
		{
			f[i][0][0] = f[i-1][0][0] + 1;;
		}
		else
		{
			f[i][0][0] = 1;
		}
	}
	for (int i = 2 ; i <= m ; i++)
	{
		if (y[i] != y[i-1])
		{
			f[0][i][1] = f[0][i-1][1] + 1;
		}
		else
		{
			f[0][i][1] = 1;
		}
	}
	for (int i = 1 ; i <= n ; i++)
	{
		for (int e = 1 ; e <= m ; e++)
		{
			if (x[i] != x[i-1])
			{
				f[i][e][0] += f[i-1][e][0];
			}
			if (x[i] != y[e])
			{
				f[i][e][0] += f[i-1][e][1];
			}
			if (y[e] != x[i])
			{
				f[i][e][1] += f[i][e-1][0];
			}
			if (y[e] != y[e-1])
			{
				f[i][e][1] += f[i][e-1][1];
			}
			if (e != 1 && x[i] != y[e])
			{
				f[i][e][1] += f[i][0][0];
			}
			if (i != 1 && x[i] != y[e])
			{
				f[i][e][0]+=f[0][e][1];
			}
			f[i][e][0] %= mod;
			f[i][e][1] %= mod;
			res += f[i][e][0];
			res += f[i][e][1];
			res %= mod;
		//	cout <<i<<" "<<e<<" "<<f[i][e][0]<<" "<<f[i][e][1]<<'\n';
		}
	}
	cout <<res<<'\n';
}



### 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 <time.h>
#include <iomanip>
#include <iostream>
#include <map>
#include <algorithm>
#include <queue>
#include <set>
#include <vector>
#include <iomanip>
#include <bitset>
using namespace std;
long long kq = 0 , f[1005][1005][3] , mod = 998244353;
string x , y;
void trau(int sti , int ste , int i , int e , string s)
{
	if (s.size() > 1)
	{
		if (s[s.size() - 1] == s[s.size() - 2])
		{
			return;
		}
		if (i != sti && e != ste)
		{
			cout <<sti<<" "<<ste<<" "<<i<<" "<<e<<" "<<s<<'\n';
			kq++;
		}
	}
	if (i+1 < x.size())
	{
		s += x[i+1];
		trau(sti , ste , i + 1 , e , s);
		s.pop_back();
	}
	if (e+1 < y.size())
	{
		s += y[e+1];
		trau(sti , ste , i , e + 1 , s);
		s.pop_back();
	}
}
int tt()
{
	for (int i = 1 ; i < x.size() ; i++)
	{
		for (int e = 1 ; e < y.size() ; e++)
		{
			trau(i-1 , e-1 , i-1 , e-1 , "");
		}
	}
	return kq;
}
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	srand (time(NULL));
	int test = 1;
	cin >> x;
	cin >> y;
	int n = x.size();
	int m = y.size();
	x = '.' + x;
	y = '/' + y;
	//cout <<tt()<<'\n';
	f[1][0][0] = 1;
	f[0][1][1] = 1;
	long long res = 0;
	for (int i = 2 ; i <= n ; i++)
	{
		if (x[i] != x[i-1])
		{
			f[i][0][0] = f[i-1][0][0] + 1;;
		}
		else
		{
			f[i][0][0] = 1;
		}
	}
	for (int i = 2 ; i <= m ; i++)
	{
		if (y[i] != y[i-1])
		{
			f[0][i][1] = f[0][i-1][1] + 1;
		}
		else
		{
			f[0][i][1] = 1;
		}
	}
	for (int i = 1 ; i <= n ; i++)
	{
		for (int e = 1 ; e <= m ; e++)
		{
			if (x[i] != x[i-1])
			{
				f[i][e][0] += f[i-1][e][0];
			}
			if (x[i] != y[e])
			{
				f[i][e][0] += f[i-1][e][1];
			}
			if (y[e] != x[i])
			{
				f[i][e][1] += f[i][e-1][0];
			}
			if (y[e] != y[e-1])
			{
				f[i][e][1] += f[i][e-1][1];
			}
			if (e != 1 && x[i] != y[e])
			{
				f[i][e][1] += f[i][0][0];
			}
			if (i != 1 && x[i] != y[e])
			{
				f[i][e][0]+=f[0][e][1];
			}
			f[i][e][0] %= mod;
			f[i][e][1] %= mod;
			res += f[i][e][0];
			res += f[i][e][1];
			res %= mod;
		//	cout <<i<<" "<<e<<" "<<f[i][e][0]<<" "<<f[i][e][1]<<'\n';
		}
	}
	cout <<res<<'\n';
}



```