#include<bits/stdc++.h>

using namespace std;
#define show(x) cout<< #x << "= " << x << " \n" ;
#define mod 998244353
#define INF 1000000000000000000
#define pie 3.14159265358979323846
#define ll long long int
#define pb push_back
#define popcnt __builtin_popcount
#define mp make_pair
#define pii pair<ll,ll>
#define prod(a,b) ((a*b)%mod)
#define set0(x) memset(x,0,sizeof(x))
#define setm1(x) memset(x,-1,sizeof(x))
#define printlist(arr) for(auto lstelem :arr) cout<< lstelem << " ";cout<<" \n" 
#define all(a) a.begin(),a.end()
#define sqre(a) ((a)*(a))

//maths stuff
ll powr(ll a,ll b){
    ll res=1;
    while(b){
        if(b%2)
            res=(a*res)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return res;
}

void extentded_eucledian(ll a,ll b,ll &ex,ll &ey){
    if(b==0){
        ex=1;
        ey=0;
        return;
    }
    extentded_eucledian(b,a%b,ex,ey);
    swap(ex,ey);
    ey-= (a/b)*ex;//gcd= x*a+y*b
}

ll dp[1002][1002][2];

int main(){ 
    #ifdef ONLINE_JUDGE
        //onlinesubmission
    #else
    freopen("input.txt","r",stdin);
    #endif
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    string x,y; cin>> x >> y ;
    ll n=x.size(),m=y.size();
    set0(dp);
    ll ans=0;
    for(ll i=0;i<=n;++i){
        for(ll j=0;j<=m;++j){
            if(i>0) dp[i][j][0]=(dp[i][j][0]+1)%mod;
            if(j>0) dp[i][j][1]=(dp[i][j][1]+1)%mod;
            if(i>1){
                if(x[i-1]!=x[i-2]) dp[i][j][0]=(dp[i][j][0]+dp[i-1][j][0])%mod;
            }
            if(i>0 && j>0){
                if(x[i-1]!=y[j-1]){
                    dp[i][j][0]=(dp[i][j][0]+dp[i-1][j][1])%mod;
                    dp[i][j][1]=(dp[i][j][1]+dp[i][j-1][0])%mod;
                }
            }
            if(j>1){
                if(y[j-1]!=y[j-2]) dp[i][j][1]=(dp[i][j][1]+dp[i][j-1][1])%mod;
            }
            ans=(ans+dp[i][j][0]+dp[i][j][1])%mod;
        }
    }
    for(ll i=0;i<n;++i){
        ll j=i;
        while(j+1<n) if(x[j]!=x[j+1]) ++j; else break;
        ll l=(j-i+1),v=(( ((l+1)*l/2)%mod )*(m+1))%mod;
        ans=(ans-v+mod)%mod;
        i=j;
    }
    for(ll i=0;i<m;++i){
        ll j=i;
        while(j+1<m) if(y[j]!=y[j+1]) ++j; else break;
        ll l=(j-i+1),v=((((l+1)*l/2)%mod)*(n+1))%mod;
        ans=(ans-v+mod)%mod;
        i=j;
    }
    cout<< ans ;
    return 0;
}

//might1331 pratapSingh:"Try Kr raha hu mai bc, timer laga rakha h"
### Prompt

In cpp, your task is to solve 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>

using namespace std;
#define show(x) cout<< #x << "= " << x << " \n" ;
#define mod 998244353
#define INF 1000000000000000000
#define pie 3.14159265358979323846
#define ll long long int
#define pb push_back
#define popcnt __builtin_popcount
#define mp make_pair
#define pii pair<ll,ll>
#define prod(a,b) ((a*b)%mod)
#define set0(x) memset(x,0,sizeof(x))
#define setm1(x) memset(x,-1,sizeof(x))
#define printlist(arr) for(auto lstelem :arr) cout<< lstelem << " ";cout<<" \n" 
#define all(a) a.begin(),a.end()
#define sqre(a) ((a)*(a))

//maths stuff
ll powr(ll a,ll b){
    ll res=1;
    while(b){
        if(b%2)
            res=(a*res)%mod;
        a=(a*a)%mod;
        b/=2;
    }
    return res;
}

void extentded_eucledian(ll a,ll b,ll &ex,ll &ey){
    if(b==0){
        ex=1;
        ey=0;
        return;
    }
    extentded_eucledian(b,a%b,ex,ey);
    swap(ex,ey);
    ey-= (a/b)*ex;//gcd= x*a+y*b
}

ll dp[1002][1002][2];

int main(){ 
    #ifdef ONLINE_JUDGE
        //onlinesubmission
    #else
    freopen("input.txt","r",stdin);
    #endif
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    string x,y; cin>> x >> y ;
    ll n=x.size(),m=y.size();
    set0(dp);
    ll ans=0;
    for(ll i=0;i<=n;++i){
        for(ll j=0;j<=m;++j){
            if(i>0) dp[i][j][0]=(dp[i][j][0]+1)%mod;
            if(j>0) dp[i][j][1]=(dp[i][j][1]+1)%mod;
            if(i>1){
                if(x[i-1]!=x[i-2]) dp[i][j][0]=(dp[i][j][0]+dp[i-1][j][0])%mod;
            }
            if(i>0 && j>0){
                if(x[i-1]!=y[j-1]){
                    dp[i][j][0]=(dp[i][j][0]+dp[i-1][j][1])%mod;
                    dp[i][j][1]=(dp[i][j][1]+dp[i][j-1][0])%mod;
                }
            }
            if(j>1){
                if(y[j-1]!=y[j-2]) dp[i][j][1]=(dp[i][j][1]+dp[i][j-1][1])%mod;
            }
            ans=(ans+dp[i][j][0]+dp[i][j][1])%mod;
        }
    }
    for(ll i=0;i<n;++i){
        ll j=i;
        while(j+1<n) if(x[j]!=x[j+1]) ++j; else break;
        ll l=(j-i+1),v=(( ((l+1)*l/2)%mod )*(m+1))%mod;
        ans=(ans-v+mod)%mod;
        i=j;
    }
    for(ll i=0;i<m;++i){
        ll j=i;
        while(j+1<m) if(y[j]!=y[j+1]) ++j; else break;
        ll l=(j-i+1),v=((((l+1)*l/2)%mod)*(n+1))%mod;
        ans=(ans-v+mod)%mod;
        i=j;
    }
    cout<< ans ;
    return 0;
}

//might1331 pratapSingh:"Try Kr raha hu mai bc, timer laga rakha h"
```