#include <bits/stdc++.h>
using namespace std;
#define mem(x) memset(x, 0, sizeof x)
#define all(x) (x).begin(), (x).end()
#define endl "\n"
typedef long long ll;
inline ll rint() { ll x = 0, f = 1; char c = getchar(); while (c<'0' || c>'9') { if (c == '-')f = -1; c = getchar(); }while (c >= '0'&&c <= '9') { x = x * 10 + c - '0'; c = getchar(); }return x * f; }
inline ll gcd(ll a, ll b) { return (b) ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }
inline ll pow(ll a, ll b, ll q) { ll rtn = 1; while (b) { if (b & 1) rtn = rtn * a%q; a = a * a%q; b >>= 1; } return rtn; }
inline ll mysqrt(ll n) { ll x = sqrt(n); while (x*x < n) x++; while (x*x > n) x--; return x; }

const ll mod = 998244353;
char x[1005], y[1005];
ll dp[1005][1005][2];
ll dpx[1005];
ll dpy[1005];

inline void upd(ll &a, ll b) {a+=b;a%=mod;}
void solve() {
  scanf("%s", x+1);
  scanf("%s", y+1);
  int n=strlen(x+1);
  int m=strlen(y+1);
  ll res=0;
  for(int i=1;i<=m;i++) {
    dpy[i]=1;
    if(y[i-1]!=y[i]) upd(dpy[i],dpy[i-1]);
  }
  for(int i=1;i<=n;i++) {
    dpx[i]=1;
    if (x[i-1]!=x[i]) upd(dpx[i], dpx[i-1]);
    for (int j=1;j<=m;j++) {
      if(x[i]!=y[j]) upd(dp[i][j][0], dpy[j]);
      if(x[i]!=y[j]) upd(dp[i][j][1], dpx[i]);
      if(x[i-1]!=x[i]) upd(dp[i][j][0], dp[i-1][j][0]);
      if(y[j]!=x[i]) upd(dp[i][j][0], dp[i-1][j][1]);
      if(x[i]!=y[j]) upd(dp[i][j][1], dp[i][j-1][0]);
      if(y[j-1]!=y[j]) upd(dp[i][j][1], dp[i][j-1][1]);
      upd(res, dp[i][j][0]+dp[i][j][1]);
    }
  }
  cout<<res<<endl;
}
int main() {
  int t=1;
//  t=rint();
  while(t--) solve();
  return 0;
}
### Prompt

Your challenge is to write 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 mem(x) memset(x, 0, sizeof x)
#define all(x) (x).begin(), (x).end()
#define endl "\n"
typedef long long ll;
inline ll rint() { ll x = 0, f = 1; char c = getchar(); while (c<'0' || c>'9') { if (c == '-')f = -1; c = getchar(); }while (c >= '0'&&c <= '9') { x = x * 10 + c - '0'; c = getchar(); }return x * f; }
inline ll gcd(ll a, ll b) { return (b) ? gcd(b, a%b) : a; }
inline ll lcm(ll a, ll b) { return a / gcd(a, b)*b; }
inline ll pow(ll a, ll b, ll q) { ll rtn = 1; while (b) { if (b & 1) rtn = rtn * a%q; a = a * a%q; b >>= 1; } return rtn; }
inline ll mysqrt(ll n) { ll x = sqrt(n); while (x*x < n) x++; while (x*x > n) x--; return x; }

const ll mod = 998244353;
char x[1005], y[1005];
ll dp[1005][1005][2];
ll dpx[1005];
ll dpy[1005];

inline void upd(ll &a, ll b) {a+=b;a%=mod;}
void solve() {
  scanf("%s", x+1);
  scanf("%s", y+1);
  int n=strlen(x+1);
  int m=strlen(y+1);
  ll res=0;
  for(int i=1;i<=m;i++) {
    dpy[i]=1;
    if(y[i-1]!=y[i]) upd(dpy[i],dpy[i-1]);
  }
  for(int i=1;i<=n;i++) {
    dpx[i]=1;
    if (x[i-1]!=x[i]) upd(dpx[i], dpx[i-1]);
    for (int j=1;j<=m;j++) {
      if(x[i]!=y[j]) upd(dp[i][j][0], dpy[j]);
      if(x[i]!=y[j]) upd(dp[i][j][1], dpx[i]);
      if(x[i-1]!=x[i]) upd(dp[i][j][0], dp[i-1][j][0]);
      if(y[j]!=x[i]) upd(dp[i][j][0], dp[i-1][j][1]);
      if(x[i]!=y[j]) upd(dp[i][j][1], dp[i][j-1][0]);
      if(y[j-1]!=y[j]) upd(dp[i][j][1], dp[i][j-1][1]);
      upd(res, dp[i][j][0]+dp[i][j][1]);
    }
  }
  cout<<res<<endl;
}
int main() {
  int t=1;
//  t=rint();
  while(t--) solve();
  return 0;
}
```