#include <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm> 
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include  <ctime>
#include <iomanip>
#include <bitset>
#include <assert.h>   
using namespace std;
 
#define int                  long long int
#define uint                  unsigned long long int
#define loop(i,a,b)           for(int i=(int)a;i<(int)b;++i)
#define rloop(i,a,b)          for(int i=(int)a;i<=(int)b;++i)
#define loopl(i,a,b)          for(int i=(int)a;i<(int)b;++i)
#define loopr(i,a,b)          for(int i=(int)a;i>=(int)b;--i)
#define count_1(n)            __builtin_popcountint(n)
#define pb                    push_back
#define eb                    emplace_back
#define ab(a)                 (a<0)?(-1*a):a
#define pc                    putchar
#define gc                    getchar
#define mset(a,b,c)           loop(i,0,b) a[i]=c
#define mp                    make_pair
#define itoc(c)               ((char)(((int)'0')+c))
#define aint(p)                p.begin(),p.end()
#define max(x,y)              ((x>y)?x:y)
#define min(x,y)              ((x<y)?x:y)
#define mid(s,e)              (s+(e-s)/2)
#define vi                    vector<int>
#define vvi                   vector<vector<int>>
 
const int MOD = 998244353;
 
int modpow(int a, int b, int mod) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; res %= mod; a = a * a; a %= mod; }return res; }
int p(int a, int b) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; a = a * a; }return res; }


void testcases() {
    string a,b; cin>>a>>b;
    int n = a.size(); int m = b.size();
    a = '$'+a; b = '$'+b;
    vector<vvi> dp(n+1, vvi(m+1, vi(2)));
    loop(i,1,n+1) { dp[i][0][0]++; dp[i][0][0]%=MOD; if(a[i]!=a[i-1]) { dp[i][0][0] += dp[i-1][0][0]; dp[i][0][0]%=MOD; } }
    loop(j,1,m+1) { dp[0][j][1]++; dp[0][j][1]%=MOD; if(b[j]!=b[j-1]) { dp[0][j][1] += dp[0][j-1][1]; dp[0][j][1]%=MOD; } }
    int ans = 0;
    loop(i,1,n+1) {
        loop(j,1,m+1) {
            dp[i][j][0]++; dp[i][j][1]++;
            if(a[i]!=a[i-1]) { dp[i][j][0] += dp[i-1][j][0]; dp[i][j][0]%=MOD; }
            if(a[i]!=b[j]) { dp[i][j][0] += dp[i-1][j][1]; dp[i][j][0]%=MOD; }
            if(b[j]!=a[i]) { dp[i][j][1] += dp[i][j-1][0]; dp[i][j][1]%=MOD; }
            if(b[j]!=b[j-1]) { dp[i][j][1] += dp[i][j-1][1]; dp[i][j][1]%=MOD; }
            ans += dp[i][j][0]; ans%=MOD;
            ans += dp[i][j][1]; ans%=MOD;
        }
    }
    int xans = 0; int yans = 0; int xpos = 1; int ypos = 1;
    loop(i,2,n+1) if(a[i]==a[i-1]) { xans += (((i-xpos)*(i-xpos+1))%MOD * modpow(2,MOD-2,MOD))%MOD; xans%=MOD; xpos = i; } xans += (((n+1-xpos)*(n+1-xpos+1))%MOD * modpow(2,MOD-2,MOD))%MOD; xans%=MOD;
    loop(i,2,m+1) if(b[i]==b[i-1]) { yans += (((i-ypos)*(i-ypos+1))%MOD * modpow(2,MOD-2,MOD))%MOD; ypos = i; } yans += (((m+1-ypos)*(m+1-ypos+1))%MOD * modpow(2,MOD-2,MOD))%MOD;

    ans -= (xans*(m))%MOD; ans+=MOD; ans%=MOD;
    ans -= (yans*(n))%MOD; ans+=MOD; ans%=MOD;

    cout << ans << "\n";
}     
 
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    /* code */ 
    int t; 
    // cin >> t;
    t = 1;
    loop(i,0,t) testcases();
    /* end code */
}
### 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 <iostream>
#include <sstream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cctype>
#include <string>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm> 
#include <functional>
#include <unordered_map>
#include <unordered_set>
#include  <ctime>
#include <iomanip>
#include <bitset>
#include <assert.h>   
using namespace std;
 
#define int                  long long int
#define uint                  unsigned long long int
#define loop(i,a,b)           for(int i=(int)a;i<(int)b;++i)
#define rloop(i,a,b)          for(int i=(int)a;i<=(int)b;++i)
#define loopl(i,a,b)          for(int i=(int)a;i<(int)b;++i)
#define loopr(i,a,b)          for(int i=(int)a;i>=(int)b;--i)
#define count_1(n)            __builtin_popcountint(n)
#define pb                    push_back
#define eb                    emplace_back
#define ab(a)                 (a<0)?(-1*a):a
#define pc                    putchar
#define gc                    getchar
#define mset(a,b,c)           loop(i,0,b) a[i]=c
#define mp                    make_pair
#define itoc(c)               ((char)(((int)'0')+c))
#define aint(p)                p.begin(),p.end()
#define max(x,y)              ((x>y)?x:y)
#define min(x,y)              ((x<y)?x:y)
#define mid(s,e)              (s+(e-s)/2)
#define vi                    vector<int>
#define vvi                   vector<vector<int>>
 
const int MOD = 998244353;
 
int modpow(int a, int b, int mod) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; res %= mod; a = a * a; a %= mod; }return res; }
int p(int a, int b) { int res = 1; for (; b; b >>= 1) { if (b & 1)res = res * a; a = a * a; }return res; }


void testcases() {
    string a,b; cin>>a>>b;
    int n = a.size(); int m = b.size();
    a = '$'+a; b = '$'+b;
    vector<vvi> dp(n+1, vvi(m+1, vi(2)));
    loop(i,1,n+1) { dp[i][0][0]++; dp[i][0][0]%=MOD; if(a[i]!=a[i-1]) { dp[i][0][0] += dp[i-1][0][0]; dp[i][0][0]%=MOD; } }
    loop(j,1,m+1) { dp[0][j][1]++; dp[0][j][1]%=MOD; if(b[j]!=b[j-1]) { dp[0][j][1] += dp[0][j-1][1]; dp[0][j][1]%=MOD; } }
    int ans = 0;
    loop(i,1,n+1) {
        loop(j,1,m+1) {
            dp[i][j][0]++; dp[i][j][1]++;
            if(a[i]!=a[i-1]) { dp[i][j][0] += dp[i-1][j][0]; dp[i][j][0]%=MOD; }
            if(a[i]!=b[j]) { dp[i][j][0] += dp[i-1][j][1]; dp[i][j][0]%=MOD; }
            if(b[j]!=a[i]) { dp[i][j][1] += dp[i][j-1][0]; dp[i][j][1]%=MOD; }
            if(b[j]!=b[j-1]) { dp[i][j][1] += dp[i][j-1][1]; dp[i][j][1]%=MOD; }
            ans += dp[i][j][0]; ans%=MOD;
            ans += dp[i][j][1]; ans%=MOD;
        }
    }
    int xans = 0; int yans = 0; int xpos = 1; int ypos = 1;
    loop(i,2,n+1) if(a[i]==a[i-1]) { xans += (((i-xpos)*(i-xpos+1))%MOD * modpow(2,MOD-2,MOD))%MOD; xans%=MOD; xpos = i; } xans += (((n+1-xpos)*(n+1-xpos+1))%MOD * modpow(2,MOD-2,MOD))%MOD; xans%=MOD;
    loop(i,2,m+1) if(b[i]==b[i-1]) { yans += (((i-ypos)*(i-ypos+1))%MOD * modpow(2,MOD-2,MOD))%MOD; ypos = i; } yans += (((m+1-ypos)*(m+1-ypos+1))%MOD * modpow(2,MOD-2,MOD))%MOD;

    ans -= (xans*(m))%MOD; ans+=MOD; ans%=MOD;
    ans -= (yans*(n))%MOD; ans+=MOD; ans%=MOD;

    cout << ans << "\n";
}     
 
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    /* code */ 
    int t; 
    // cin >> t;
    t = 1;
    loop(i,0,t) testcases();
    /* end code */
}
```