#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
#define owo(i,a, b) for(int i=(a);i<(b); ++i)
#define uwu(i,a, b) for(int i=(a)-1; i>=(b); --i)
#define senpai push_back
#define ttgl pair<int, int>
#define ayaya cout<<"ayaya~"<<endl
 
using namespace std;
using ll = long long;
using ld = long double;
ll MOD = 998244353;
const ll root = 3;
ll binpow(ll a,ll b){ll res=1;while(b){if(b&1)res=(res*a)%MOD;a=(a*a)%MOD;b>>=1;}return res;}
ll modInv(ll a){return binpow(a, MOD-2);}
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 1001;
ll dp[mxN][mxN][4];
string x, y;
int main() {
    //freopen("file.in", "r", stdin);
    //freopen("file.out", "w", stdout);
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    cin.tie(0)->sync_with_stdio(0);
    string s;
    cin>>s;
    x = "?" + s;
    cin>>s;
    y = "?" + s;
    ll ans = 0;
    owo(i, 0, x.size()) {
        owo(j, 0, y.size()) {
            //add one from x
            if(i > 0) {
                dp[i][j][2] = 1;
                if(x[i]!=x[i-1]) {
                    dp[i][j][0] = (dp[i][j][0] + dp[i-1][j][0])%MOD;
                    dp[i][j][2] = (dp[i][j][2] + dp[i-1][j][2])%MOD;
                }
                if(x[i]!=y[j]) {
                    dp[i][j][0] = (dp[i][j][0] + dp[i-1][j][1])%MOD;
                    dp[i][j][0] = (dp[i][j][0] + dp[i-1][j][3])%MOD;
                }
            }
            if(j > 0) {
                dp[i][j][3] = 1;
                if(y[j]!=y[j-1]) {
                    dp[i][j][1] = (dp[i][j][1] + dp[i][j-1][1])%MOD;
                    dp[i][j][3] = (dp[i][j][3] + dp[i][j-1][3])%MOD;
                }
                if(x[i]!=y[j]) {
                    dp[i][j][1] = (dp[i][j][1] + dp[i][j-1][0])%MOD;
                    dp[i][j][1] = (dp[i][j][1] + dp[i][j-1][2])%MOD;
                }
            }
            ans = (ans + dp[i][j][0] + dp[i][j][1])%MOD;
            //cout<<i<<" "<<j<<" "<<dp[i][j][0]<<" "<<dp[i][j][1]<<" "<<dp[i][j][2]<<" "<<dp[i][j][3]<<" "<<x[i]<<" "<<y[j]<<"\n";
        }
    }
    cout<<ans<<"\n";
    return 0;
}
### Prompt

Please formulate 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
#pragma GCC target ("avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
#define owo(i,a, b) for(int i=(a);i<(b); ++i)
#define uwu(i,a, b) for(int i=(a)-1; i>=(b); --i)
#define senpai push_back
#define ttgl pair<int, int>
#define ayaya cout<<"ayaya~"<<endl
 
using namespace std;
using ll = long long;
using ld = long double;
ll MOD = 998244353;
const ll root = 3;
ll binpow(ll a,ll b){ll res=1;while(b){if(b&1)res=(res*a)%MOD;a=(a*a)%MOD;b>>=1;}return res;}
ll modInv(ll a){return binpow(a, MOD-2);}
const int INF = 0x3f3f3f3f;
const int NINF = 0xc0c0c0c0;
const ll INFLL = 0x3f3f3f3f3f3f3f3f;
const ll NINFLL = 0xc0c0c0c0c0c0c0c0;
const int mxN = 1001;
ll dp[mxN][mxN][4];
string x, y;
int main() {
    //freopen("file.in", "r", stdin);
    //freopen("file.out", "w", stdout);
    mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
    cin.tie(0)->sync_with_stdio(0);
    string s;
    cin>>s;
    x = "?" + s;
    cin>>s;
    y = "?" + s;
    ll ans = 0;
    owo(i, 0, x.size()) {
        owo(j, 0, y.size()) {
            //add one from x
            if(i > 0) {
                dp[i][j][2] = 1;
                if(x[i]!=x[i-1]) {
                    dp[i][j][0] = (dp[i][j][0] + dp[i-1][j][0])%MOD;
                    dp[i][j][2] = (dp[i][j][2] + dp[i-1][j][2])%MOD;
                }
                if(x[i]!=y[j]) {
                    dp[i][j][0] = (dp[i][j][0] + dp[i-1][j][1])%MOD;
                    dp[i][j][0] = (dp[i][j][0] + dp[i-1][j][3])%MOD;
                }
            }
            if(j > 0) {
                dp[i][j][3] = 1;
                if(y[j]!=y[j-1]) {
                    dp[i][j][1] = (dp[i][j][1] + dp[i][j-1][1])%MOD;
                    dp[i][j][3] = (dp[i][j][3] + dp[i][j-1][3])%MOD;
                }
                if(x[i]!=y[j]) {
                    dp[i][j][1] = (dp[i][j][1] + dp[i][j-1][0])%MOD;
                    dp[i][j][1] = (dp[i][j][1] + dp[i][j-1][2])%MOD;
                }
            }
            ans = (ans + dp[i][j][0] + dp[i][j][1])%MOD;
            //cout<<i<<" "<<j<<" "<<dp[i][j][0]<<" "<<dp[i][j][1]<<" "<<dp[i][j][2]<<" "<<dp[i][j][3]<<" "<<x[i]<<" "<<y[j]<<"\n";
        }
    }
    cout<<ans<<"\n";
    return 0;
}
```