using namespace std;

#include <set>
#include <map>
#include <list>
#include <deque>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <vector>
#include <string>
#include <bitset>
#include <utility>
#include <algorithm>

#define pb push_back
#define sz size
#define f first
#define s second
#define II inline
#define ll long long
#define db double
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define all(v) v.begin() , v.end()
#define CC(v) memset((v),0,sizeof((v)))
#define CP(v,w) memcpy((v),(w),sizeof((w)))
#define mp make_pair

#define IN "iv.in"
#define OUT "iv.out"
#define N_MAX 1<<9
#define mod 3210121

typedef vector<int> VI;
typedef pair<int,int> pi;
typedef vector<string> VS;

int rez,N,M;
char V1[N_MAX],V2[N_MAX];
int C[N_MAX][N_MAX],B[N_MAX][N_MAX];
int h1,h2,h3,h4;
int F[1<<8];
int poz1,poz2;

II void scan()
{
	freopen(IN,"r",stdin);
	freopen(OUT,"w",stdout);
	fgets(V1+1,1<<10,stdin);
	fgets(V2+1,1<<10,stdin);
	
	N = strlen(V1+1) - 1;
	M = strlen(V2+1) - 1;
	if(V2[M+1] != '\0' && V2[M+1] != '\n')
		++M;
	
	FOR(i,1,N)
		++F[ V1[i] ];
	FOR(i,1,M)
		++F[ V2[i] ];
	FOR(i,'a','z')
	if(F[i] == 1)
	{
		FOR(j,1,N)
			if(V1[j] == i)
				poz1 = j;
		FOR(j,1,M)
			if(V2[j] == i)
				poz2 = j;
	}	
}

II void make(int i1,int C[N_MAX][N_MAX],int B[N_MAX][N_MAX])
{
	for(int i2=N+1;i2>i1;--i2)
	FOR(i3,0,M)
	{
		int i4 = (N+M+2) - (i3+i1+i2);
		if(!i4)
			break;
		for(;B[i2][i3] >= mod;B[i2][i3] -= mod);
	
		if(V1[i1 + 1] == V1[i2 - 1])
			C[i2-1][i3] += B[i2][i3];
		if(i1!=N && i4!=1 && V1[i1 + 1] == V2[i4 - 1])
			C[i2][i3] += B[i2][i3];
		if(V2[i3 + 1] == V1[i2 - 1])
			B[i2-1][i3+1] += B[i2][i3];
		if(V2[i3 + 1] == V2[i4 - 1])
			B[i2][i3+1] += B[i2][i3];	
		
	//	printf("%d %d %d %d : %d\n",i1,i2,i3,i4,B[i2][i3]);
		
		if(!poz1 && !poz2 && i1+1 == i2 && i3+1 == i4)
		{
			rez += B[i2][i3];	
			for(;rez >= mod;rez -= mod);
		}
		if(i1==1 && i2==3 && i3==1 && i4==2)
		{
			int debug = 1;
		}
		if(poz1 && i1==poz1-1 && i2==poz1+1 && i3+1==i4)
		{
			rez += B[i2][i3];	
			for(;rez >= mod;rez -= mod);
		}
		if(poz2 && i1+1==i2 && i3==poz2-1 && i4==poz2+1)
		{
			rez += B[i2][i3];	
			for(;rez >= mod;rez -= mod);
		}	
		
	}
}

II void solve()
{
	B[N+1][0] = 1;
	
	
	FOR(i1,0,N)
		if(i1&1)
		{
			CC(B);
			make(i1,B,C);
		}	
		else
		{
			CC(C);	
			make(i1,C,B);
		}	
	printf("%d\n",rez);
}

int main()
{
	scan();
	solve();
	return 0;
}