#include <bits/stdc++.h>

// #pragma GCC target("sse,sse2,sse3,ssse3,sse4")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;

#define fst first
#define snd second
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back

#define all(v) (v).begin(), (v).end()
#define sz(v) ((int)(v).size())
#define sqr(x) ((x) * (x))

typedef long long ll;
typedef  double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef vector<int> vi;

#define next ajksdslk
#define prev aklsfjk

mt19937_64 mt_rand(chrono::system_clock::now().time_since_epoch().count());

template<typename T1, typename T2> inline bool upmax(T1& a, T2 b) { return (a < b ? (a = b, true) : false); }
template<typename T1, typename T2> inline bool upmin(T1& a, T2 b) { return (b < a ? (a = b, true) : false); }

const int maxn = (int) 5e5 + 100;
const int maxlog = 21;
const int base = (int) 1e9 + 7;
const ld eps = (ld) 1e-9;
const ld PI = acos(-1.);
//const int pp = 41;
const int inf = 2e9;
const ll llinf = 1e18;

int f[1111][1111][2];
const int mod = 998244353;

void add(int & a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}

void solve() {

  string a, b;
  cin >> a >> b;

  vector<int> fa(sz(a) + 1, 0), fb(sz(b) + 1, 0);
  fa[0] = fb[0] = 1;
  for (int i = 0; i < sz(a); i ++) {
    if (i && a[i - 1] != a[i]) add(fa[i + 1], fa[i]);
    add(fa[i + 1], 1);

  }
  for (int i = 0; i < sz(b); i ++) {
    if (i && b[i - 1] != b[i]) add(fb[i + 1], fb[i]);
    add(fb[i + 1], 1);

  }
//  for (int i = 0; i <= sz(a); i ++) {
//    cout << fa[i] << " ";
////    f[i][0][0] = fa[i];
//  }
//  cout << '\n';
//  for (int i = 0; i <= sz(b); i ++) {
//    cout << fb[i] << " ";
////    f[0][i][1] = fb[i];
//  }
//  cout << "\n";
//  cout << "\n";
////

  for (int prefA = 0; prefA <= sz(a); prefA ++) {
    for (int prefB = 0; prefB <= sz(b); prefB ++) {
      if (prefA + prefB == 0) continue;
      // case A
      if (prefA < sz(a) && prefB) {
        bool eqA = prefA ? ( a[prefA - 1] == a[prefA] ) : false;
        bool eqB = b[prefB - 1] == a[prefA];

        if (!eqA) {
          add(f[prefA + 1][prefB][0], f[prefA][prefB][0]);
        }
        if (!eqB) {
          add(f[prefA + 1][prefB][0], f[prefA][prefB][1]);
          add(f[prefA + 1][prefB][0], fb[prefB]);
        }

      }
      // case B
      if (prefB < sz(b) && prefA) {
        bool eqA = a[prefA - 1] == b[prefB];
        bool eqB = prefB ? b[prefB - 1] == b[prefB] : false;

        if (!eqA) {
          add(f[prefA][prefB + 1][1], f[prefA][prefB][0]);
          add(f[prefA][prefB + 1][1], fa[prefA]);
        }
        if (!eqB) {
          add(f[prefA][prefB + 1][1], f[prefA][prefB][1]);
        }


      }

    }
  }
  int ans = 0;
  for (int i = 1; i <= sz(a); i ++){
    for (int j = 1; j <= sz(b); j ++) {
//      cout << f[i][j][0] << " ";
      add(ans, f[i][j][0]);
      add(ans, f[i][j][1]);
    }
//    cout << "\n";
  }
  cout << ans << "\n";

}




int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
#ifdef LOCAL
  freopen ("input.txt", "r", stdin);
#endif


//  for (int i = 1; i < maxn; i ++) {
//    cout << getCnt(i) << " " << f[i] << "\n";
//  }


  int t = 1;
//  cin >> t;
  while (t --) {
    solve();
  }



  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>

// #pragma GCC target("sse,sse2,sse3,ssse3,sse4")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
using namespace std;

#define fst first
#define snd second
#define mp make_pair
#define mt make_tuple
#define pb push_back
#define eb emplace_back

#define all(v) (v).begin(), (v).end()
#define sz(v) ((int)(v).size())
#define sqr(x) ((x) * (x))

typedef long long ll;
typedef  double ld;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef vector<int> vi;

#define next ajksdslk
#define prev aklsfjk

mt19937_64 mt_rand(chrono::system_clock::now().time_since_epoch().count());

template<typename T1, typename T2> inline bool upmax(T1& a, T2 b) { return (a < b ? (a = b, true) : false); }
template<typename T1, typename T2> inline bool upmin(T1& a, T2 b) { return (b < a ? (a = b, true) : false); }

const int maxn = (int) 5e5 + 100;
const int maxlog = 21;
const int base = (int) 1e9 + 7;
const ld eps = (ld) 1e-9;
const ld PI = acos(-1.);
//const int pp = 41;
const int inf = 2e9;
const ll llinf = 1e18;

int f[1111][1111][2];
const int mod = 998244353;

void add(int & a, int b) {
  a += b;
  if (a >= mod) a -= mod;
}

void solve() {

  string a, b;
  cin >> a >> b;

  vector<int> fa(sz(a) + 1, 0), fb(sz(b) + 1, 0);
  fa[0] = fb[0] = 1;
  for (int i = 0; i < sz(a); i ++) {
    if (i && a[i - 1] != a[i]) add(fa[i + 1], fa[i]);
    add(fa[i + 1], 1);

  }
  for (int i = 0; i < sz(b); i ++) {
    if (i && b[i - 1] != b[i]) add(fb[i + 1], fb[i]);
    add(fb[i + 1], 1);

  }
//  for (int i = 0; i <= sz(a); i ++) {
//    cout << fa[i] << " ";
////    f[i][0][0] = fa[i];
//  }
//  cout << '\n';
//  for (int i = 0; i <= sz(b); i ++) {
//    cout << fb[i] << " ";
////    f[0][i][1] = fb[i];
//  }
//  cout << "\n";
//  cout << "\n";
////

  for (int prefA = 0; prefA <= sz(a); prefA ++) {
    for (int prefB = 0; prefB <= sz(b); prefB ++) {
      if (prefA + prefB == 0) continue;
      // case A
      if (prefA < sz(a) && prefB) {
        bool eqA = prefA ? ( a[prefA - 1] == a[prefA] ) : false;
        bool eqB = b[prefB - 1] == a[prefA];

        if (!eqA) {
          add(f[prefA + 1][prefB][0], f[prefA][prefB][0]);
        }
        if (!eqB) {
          add(f[prefA + 1][prefB][0], f[prefA][prefB][1]);
          add(f[prefA + 1][prefB][0], fb[prefB]);
        }

      }
      // case B
      if (prefB < sz(b) && prefA) {
        bool eqA = a[prefA - 1] == b[prefB];
        bool eqB = prefB ? b[prefB - 1] == b[prefB] : false;

        if (!eqA) {
          add(f[prefA][prefB + 1][1], f[prefA][prefB][0]);
          add(f[prefA][prefB + 1][1], fa[prefA]);
        }
        if (!eqB) {
          add(f[prefA][prefB + 1][1], f[prefA][prefB][1]);
        }


      }

    }
  }
  int ans = 0;
  for (int i = 1; i <= sz(a); i ++){
    for (int j = 1; j <= sz(b); j ++) {
//      cout << f[i][j][0] << " ";
      add(ans, f[i][j][0]);
      add(ans, f[i][j][1]);
    }
//    cout << "\n";
  }
  cout << ans << "\n";

}




int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
#ifdef LOCAL
  freopen ("input.txt", "r", stdin);
#endif


//  for (int i = 1; i < maxn; i ++) {
//    cout << getCnt(i) << " " << f[i] << "\n";
//  }


  int t = 1;
//  cin >> t;
  while (t --) {
    solve();
  }



  return 0;
}


```