#include <bits/stdc++.h>
#define LOG(FMT...) fprintf(stderr, FMT);
#define rep(i, j, k) for(int i = j; i <= k; ++ i)
#define per(i, j, k) for(int i = j; i >= k; -- i)
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
typedef pair<int, int> pii;
typedef double db;
typedef long long ll;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
mt19937 mt(std::chrono::system_clock::now().time_since_epoch().count());
uniform_int_distribution<ll> ran(0, 1ll << 62);
void ucin() { ios::sync_with_stdio(0); cin.tie(0); }
// uniform_real_distribution<double> dbran;
template<class T> inline void chkmax(T &x, const T &y) { if(x < y) x = y; }
template<class T> inline void chkmin(T &x, const T &y) { if(x > y) x = y; }
inline ll sqr(ll x) { return x * x; }
inline ll cub(ll x) { return x * x * x; }

const int N = 1000 + 10;
const int mod = 998244353;
int f[N][N][2], n, m, la[N], lb[N];
char a[N], b[N];
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int dec(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline void ADD(int &x, const int &y) { (x += y) >= mod ? x -= mod : 0; }
inline void DEC(int &x, const int &y) { (x -= y) < 0 ? x += mod : 0; }

int main() {
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  rep(i, 1, n) {
    la[i] = 1;
    while(la[i] < i && a[i - la[i]] != a[i - la[i] + 1]) la[i] ++;
  }
  rep(i, 1, m) {
    lb[i] = 1;
    while(lb[i] < i && b[i - lb[i]] != b[i - lb[i] + 1]) lb[i] ++;
  }
  int ans = 0;
  rep(i, 1, n) {
    rep(j, 1, m) {
      if(a[i] != b[j]) {
        ADD(f[i][j][0], lb[j]);
        ADD(f[i][j][1], la[i]);
      }
      if(i < n && a[i] != a[i + 1]) {
        ADD(f[i + 1][j][0], f[i][j][0]);
      }
      if(j < m && b[j] != b[j + 1]) {
        ADD(f[i][j + 1][1], f[i][j][1]);
      }
      if(i < n && a[i + 1] != b[j]) {
        ADD(f[i + 1][j][0], f[i][j][1]);
      }
      if(j < m && a[i] != b[j + 1]) {
        ADD(f[i][j + 1][1], f[i][j][0]);
      }
      ADD(ans, f[i][j][0]);
      ADD(ans, f[i][j][1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
### Prompt

Please create a solution in CPP 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>
#define LOG(FMT...) fprintf(stderr, FMT);
#define rep(i, j, k) for(int i = j; i <= k; ++ i)
#define per(i, j, k) for(int i = j; i >= k; -- i)
using namespace std;
#define fs first
#define sc second
#define pb push_back
#define mp make_pair
typedef pair<int, int> pii;
typedef double db;
typedef long long ll;
typedef long double ldb;
typedef unsigned int uint;
typedef unsigned long long ull;
mt19937 mt(std::chrono::system_clock::now().time_since_epoch().count());
uniform_int_distribution<ll> ran(0, 1ll << 62);
void ucin() { ios::sync_with_stdio(0); cin.tie(0); }
// uniform_real_distribution<double> dbran;
template<class T> inline void chkmax(T &x, const T &y) { if(x < y) x = y; }
template<class T> inline void chkmin(T &x, const T &y) { if(x > y) x = y; }
inline ll sqr(ll x) { return x * x; }
inline ll cub(ll x) { return x * x * x; }

const int N = 1000 + 10;
const int mod = 998244353;
int f[N][N][2], n, m, la[N], lb[N];
char a[N], b[N];
inline int add(int x, int y) { return x + y >= mod ? x + y - mod : x + y; }
inline int dec(int x, int y) { return x - y < 0 ? x - y + mod : x - y; }
inline void ADD(int &x, const int &y) { (x += y) >= mod ? x -= mod : 0; }
inline void DEC(int &x, const int &y) { (x -= y) < 0 ? x += mod : 0; }

int main() {
  scanf("%s%s", a + 1, b + 1);
  n = strlen(a + 1);
  m = strlen(b + 1);
  rep(i, 1, n) {
    la[i] = 1;
    while(la[i] < i && a[i - la[i]] != a[i - la[i] + 1]) la[i] ++;
  }
  rep(i, 1, m) {
    lb[i] = 1;
    while(lb[i] < i && b[i - lb[i]] != b[i - lb[i] + 1]) lb[i] ++;
  }
  int ans = 0;
  rep(i, 1, n) {
    rep(j, 1, m) {
      if(a[i] != b[j]) {
        ADD(f[i][j][0], lb[j]);
        ADD(f[i][j][1], la[i]);
      }
      if(i < n && a[i] != a[i + 1]) {
        ADD(f[i + 1][j][0], f[i][j][0]);
      }
      if(j < m && b[j] != b[j + 1]) {
        ADD(f[i][j + 1][1], f[i][j][1]);
      }
      if(i < n && a[i + 1] != b[j]) {
        ADD(f[i + 1][j][0], f[i][j][1]);
      }
      if(j < m && a[i] != b[j + 1]) {
        ADD(f[i][j + 1][1], f[i][j][0]);
      }
      ADD(ans, f[i][j][0]);
      ADD(ans, f[i][j][1]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
```