using namespace std;
#ifdef EZ
   #include "./ez/ez.h"
   const string FILE_NAME = "test";
#else
   #include <bits/stdc++.h>
   const string FILE_NAME = "iv";
#endif
#define mp make_pair
#define ll long long
#define pb push_back
#define fi first
#define se second
#define cin fin
#define cout fout
ifstream fin (FILE_NAME + ".in");
ofstream fout (FILE_NAME + ".out");

const int mod = 3210121;

int dp[11][11][11][11];

void ezsolve()
{
   string a, b;   cin >> a >> b;
   int n = a.size(), m = b.size();
   a = '3' + a;   b = '3' + b;

   for (int i = 1; i <= n; ++i)  dp[i][i][0][0] = 1;
   for (int i = 1; i < n; ++i)   dp[i][i+1][0][0] = (a[i] == a[i+1]);
   for (int len = 3; len <= n; ++len)
      for (int i = 1; i+len-1 <= n; ++i)
      {
         int j = i+len-1;
         dp[i][j][0][0] = dp[i+1][j-1][0][0] | (a[i] == a[j]);
      }

   for (int i = 1; i <= m; ++i)  dp[0][0][i][i] = 1;
   for (int i = 1; i < m; ++i)   dp[0][0][i][i+1] = (b[i] == b[i+1]);
   for (int len = 3; len <= m; ++len)
      for (int i = 1; i+len-1 <= m; ++i)
      {
         int j = i+len-1;
         dp[0][0][i][j] = dp[0][0][i+1][j-1] | (b[i] == b[j]);
      }

   // a1+b1
   for (int i = 1; i <= n; ++i)
      for (int j = 1; j <= m; ++j)
         dp[i][i][j][j] = 2 * (a[i] == b[j]);
   // a2+b1
   for (int i = 1; i < n; ++i)
      for (int j = 1; j <= m; ++j)
         dp[i][i+1][j][j] = (a[i] == b[j]) + (a[i] == a[i+1]) + (a[i+1] == b[j]);
   // a1+b2
   for (int i = 1; i <= n; ++i)
      for (int j = 1; j < m; ++j)
         dp[i][i][j][j+1] = (a[i] == b[j]) + (b[i] == b[i+1]) + (a[i] == b[j+1]);
   
   // a+b
   for (int len = 4; len <= n+m; ++len)
      for (int lena = 1; lena <= n; ++lena)
         int lenb = len-lena;
         for (int i1 = 1; i1 <= n; ++i1)
         {
            int i2 = i1+lena-1;
            for (int j1 = 1; j1 <= m; ++j1)
            {
               int j2 = j1+lenb-1;
               (dp[i1][i2][j1][j2] +=
                  (a[i1] == a[i2] ? dp[i1+1][i2-1][j1][j2] : 0) + // i1+i2
                  (b[j1] == b[j2] ? dp[i1][i2][j1+1][j2-1] : 0) + // j1+j2
                  (a[i1] == b[j2] ? dp[i1+1][i2][j1][j2-1] : 0) + // i1+j2
                  (b[j1] == a[i2] ? dp[i1][i2-1][j1+1][j2] : 0)   // j1+i2
               ) %= mod;
            }
         }
   
   cout << dp[1][n][1][m] << '\n';
}

int main()
{
   ezsolve();
}