#include<bits/stdc++.h>
using namespace std;

#define ll long long

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
  os<<"("<<p.first<<", "<<p.second<<")";
  return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
  os<<"{";
  for(int i = 0;i < (int)v.size(); i++){
    if(i)os<<", ";
    os<<v[i];
  }
  os<<"}";
  return os;
}

template<class T>
ostream& operator <<(ostream& os,const set<T>& v){
  os<<"{";
  for(auto it = v.begin(); it != v.end(); it++){
    if(it != v.begin())os<<", ";
    os<<*it;
  }
  os<<"}";
  return os;
}

template<class T, class V>
ostream& operator <<(ostream& os,const map<T, V>& v){
  os<<"{";
  for(auto it = v.begin(); it != v.end(); it++){
    if(it != v.begin())os<<", ";
    os<<*it;
  }
  os<<"}";
  return os;
}

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 <<"\n";
        //use cerr if u want to display at the bottom
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif


const ll N=1005,p=998244353;
ll dp[N][N][2],x1[N],x2[N];


int main()
{
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  string s1,s2;
  cin>>s1>>s2;
  ll n=s1.length(),m=s2.length();
  s1='#'+s1;
  s2='#'+s2;

  x1[1]=x2[1]=1;
  for(ll i=2;i<=n;i++)
  {
    if(s1[i-1]==s1[i]) x1[i]=1;
    else x1[i]=1+x1[i-1];
  }
  for(ll i=2;i<=m;i++)
  {
    if(s2[i-1]==s2[i]) x2[i]=1;
    else x2[i]=1+x2[i-1];
  }

  ll ans=0;
  if(s1[1]!=s2[1])
  {
    dp[1][1][0]=dp[1][1][1]=1;
    ans=2;
  }
  for(ll i=1;i<=n;i++)
  {
    for(ll j=1;j<=m;j++)
    {
      if(i==1 && j==1)  continue;

      // ending at s1[i]
      if(s1[i]!=s2[j])
      {
        dp[i][j][0]+=(x2[j]+dp[i-1][j][1]);
        dp[i][j][0]%=p;
      }
      if(s1[i]!=s1[i-1])
      {
        dp[i][j][0]+=dp[i-1][j][0];
        dp[i][j][0]%=p;
      }
      ans+=dp[i][j][0];
      ans%=p;

      // ending at s2[j]
      if(s2[j]!=s1[i])
      {
        dp[i][j][1]+=(x1[i]+dp[i][j-1][0]);
        dp[i][j][1]%=p;
      }
      if(s2[j]!=s2[j-1])
      {
        dp[i][j][1]+=dp[i][j-1][1];
        dp[i][j][1]%=p;
      }
      ans+=dp[i][j][1];
      ans%=p;
    }
  }

  cout<<ans<<"\n";
}

### Prompt

Develop a solution in cpp to the problem described below:

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 ll long long

template<class T,class U>
ostream& operator<<(ostream& os,const pair<T,U>& p){
  os<<"("<<p.first<<", "<<p.second<<")";
  return os;
}

template<class T>
ostream& operator <<(ostream& os,const vector<T>& v){
  os<<"{";
  for(int i = 0;i < (int)v.size(); i++){
    if(i)os<<", ";
    os<<v[i];
  }
  os<<"}";
  return os;
}

template<class T>
ostream& operator <<(ostream& os,const set<T>& v){
  os<<"{";
  for(auto it = v.begin(); it != v.end(); it++){
    if(it != v.begin())os<<", ";
    os<<*it;
  }
  os<<"}";
  return os;
}

template<class T, class V>
ostream& operator <<(ostream& os,const map<T, V>& v){
  os<<"{";
  for(auto it = v.begin(); it != v.end(); it++){
    if(it != v.begin())os<<", ";
    os<<*it;
  }
  os<<"}";
  return os;
}

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
    template <typename Arg1>
    void __f(const char* name, Arg1&& arg1){
        cout << name << " : " << arg1 <<"\n";
        //use cerr if u want to display at the bottom
    }
    template <typename Arg1, typename... Args>
    void __f(const char* names, Arg1&& arg1, Args&&... args){
        const char* comma = strchr(names + 1, ','); cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
    }
#else
#define trace(...)
#endif


const ll N=1005,p=998244353;
ll dp[N][N][2],x1[N],x2[N];


int main()
{
  ios::sync_with_stdio(false);
  cin.tie(nullptr);

  string s1,s2;
  cin>>s1>>s2;
  ll n=s1.length(),m=s2.length();
  s1='#'+s1;
  s2='#'+s2;

  x1[1]=x2[1]=1;
  for(ll i=2;i<=n;i++)
  {
    if(s1[i-1]==s1[i]) x1[i]=1;
    else x1[i]=1+x1[i-1];
  }
  for(ll i=2;i<=m;i++)
  {
    if(s2[i-1]==s2[i]) x2[i]=1;
    else x2[i]=1+x2[i-1];
  }

  ll ans=0;
  if(s1[1]!=s2[1])
  {
    dp[1][1][0]=dp[1][1][1]=1;
    ans=2;
  }
  for(ll i=1;i<=n;i++)
  {
    for(ll j=1;j<=m;j++)
    {
      if(i==1 && j==1)  continue;

      // ending at s1[i]
      if(s1[i]!=s2[j])
      {
        dp[i][j][0]+=(x2[j]+dp[i-1][j][1]);
        dp[i][j][0]%=p;
      }
      if(s1[i]!=s1[i-1])
      {
        dp[i][j][0]+=dp[i-1][j][0];
        dp[i][j][0]%=p;
      }
      ans+=dp[i][j][0];
      ans%=p;

      // ending at s2[j]
      if(s2[j]!=s1[i])
      {
        dp[i][j][1]+=(x1[i]+dp[i][j-1][0]);
        dp[i][j][1]%=p;
      }
      if(s2[j]!=s2[j-1])
      {
        dp[i][j][1]+=dp[i][j-1][1];
        dp[i][j][1]%=p;
      }
      ans+=dp[i][j][1];
      ans%=p;
    }
  }

  cout<<ans<<"\n";
}

```