#include <bits/stdc++.h>
using namespace std;
#define io_speed_up ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
template<typename T>void rd(T&x){int f=0,c;while(!isdigit(c=getchar()))f^=!(c^45);x=(c&15);while(isdigit(c=getchar()))x=x*10+(c&15);if(f)x=-x;} //读整型
template<typename T>void pt(T x,int c=10){if(x<0)putchar('-'),x=-x;if(x>9)pt(x/10,-1);putchar(x%10+48);if(c!=-1)putchar(c);}
template<typename T>void umax(T&x,const T&y){if(x<y)x=y;}
template<typename T>void umin(T&x,const T&y){if(x>y)x=y;}
#define for_seq(i,s) for(int i=0;i<(int)s.size();i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define for1(i,n) for(int i=1;i<=n;i++)
#define for0(i,n) for(int i=0;i<n;i++)
#define ms(a,b)  memset(a,b,sizeof a)
#define all(n) (n).begin(), (n).end()
#define sz(x) (int)x.size()
#define fi first
#define se second
using ll = long long;
using ld = long double;
using pii = pair<int,int>;
using pll = pair<long,long>;
const int inf = 0x3f3f3f3f;
const int maxn = 1e3+5;
const int mod = 998244353;

int dp[maxn][maxn][2];
void add(int &x,int y) {
	x += y;
	if(x>=mod) x -= mod;
}
char s[maxn],t[maxn];
int f[maxn],g[maxn];
int main()
{
	io_speed_up;
	cin>>(s+1)>>(t+1);
	int n = strlen(s+1), m = strlen(t+1);
	rep(i,1,n) f[i] = 1 + (s[i]!=s[i-1]?f[i-1]:0);
	rep(i,1,m) g[i] = 1 + (t[i]!=t[i-1]?g[i-1]:0);
	int ans = 0;
	rep(i,1,n) {
		rep(j,1,m) {
			int &x = dp[i][j][0], &y = dp[i][j][1];
			if(s[i]!=s[i-1]) {
				add(x, dp[i-1][j][0]);
			}
			if(s[i]!=t[j]) {
				add(x, dp[i-1][j][1]);
				add(x, g[j]);

				add(y, dp[i][j-1][0]);
				add(y, f[i]);
			}
			if(t[j]!=t[j-1]) {
				add(y, dp[i][j-1][1]);
			}
			add(ans, x);
			add(ans, y);
		}
	}
	cout<<ans<<"\n";
	return 0;
}
### Prompt

Please formulate 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>
using namespace std;
#define io_speed_up ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
template<typename T>void rd(T&x){int f=0,c;while(!isdigit(c=getchar()))f^=!(c^45);x=(c&15);while(isdigit(c=getchar()))x=x*10+(c&15);if(f)x=-x;} //读整型
template<typename T>void pt(T x,int c=10){if(x<0)putchar('-'),x=-x;if(x>9)pt(x/10,-1);putchar(x%10+48);if(c!=-1)putchar(c);}
template<typename T>void umax(T&x,const T&y){if(x<y)x=y;}
template<typename T>void umin(T&x,const T&y){if(x>y)x=y;}
#define for_seq(i,s) for(int i=0;i<(int)s.size();i++)
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
#define for1(i,n) for(int i=1;i<=n;i++)
#define for0(i,n) for(int i=0;i<n;i++)
#define ms(a,b)  memset(a,b,sizeof a)
#define all(n) (n).begin(), (n).end()
#define sz(x) (int)x.size()
#define fi first
#define se second
using ll = long long;
using ld = long double;
using pii = pair<int,int>;
using pll = pair<long,long>;
const int inf = 0x3f3f3f3f;
const int maxn = 1e3+5;
const int mod = 998244353;

int dp[maxn][maxn][2];
void add(int &x,int y) {
	x += y;
	if(x>=mod) x -= mod;
}
char s[maxn],t[maxn];
int f[maxn],g[maxn];
int main()
{
	io_speed_up;
	cin>>(s+1)>>(t+1);
	int n = strlen(s+1), m = strlen(t+1);
	rep(i,1,n) f[i] = 1 + (s[i]!=s[i-1]?f[i-1]:0);
	rep(i,1,m) g[i] = 1 + (t[i]!=t[i-1]?g[i-1]:0);
	int ans = 0;
	rep(i,1,n) {
		rep(j,1,m) {
			int &x = dp[i][j][0], &y = dp[i][j][1];
			if(s[i]!=s[i-1]) {
				add(x, dp[i-1][j][0]);
			}
			if(s[i]!=t[j]) {
				add(x, dp[i-1][j][1]);
				add(x, g[j]);

				add(y, dp[i][j-1][0]);
				add(y, f[i]);
			}
			if(t[j]!=t[j-1]) {
				add(y, dp[i][j-1][1]);
			}
			add(ans, x);
			add(ans, y);
		}
	}
	cout<<ans<<"\n";
	return 0;
}
```