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 N,M;
char V1[N_MAX],V2[N_MAX];
int A[N_MAX][N_MAX],B[N_MAX][N_MAX];

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;
}

II void solve()
{
	ll rez(0);
	B[0][0] = 1;
	V1[0] = V2[0] = V1[N+1] = V2[M+1] = '!';
	
	FOR(p,1,(N+M)/2)
	{
		CC(A);
		
		FOR(i2,0,p)
		FOR(i3,0,p)
		{
			int i4 = p - i3;
			int i1 = p - i2;
			if(p<<1 > N+M)
				break;
		
			if(i1 && i4 && V1[i1] == V2[M-i4+1])
				A[i2][i3] += B[i2][i3];
			if(i1 && i3 && V1[i1] == V2[i3])
				A[i2][i3] += B[i2][i3-1];
			if(i2 && i4 && V1[N-i2+1] == V2[M-i4+1])
				A[i2][i3] += B[i2-1][i3];
			if(i2 && i3 && V1[N-i2+1] == V2[i3])
				A[i2][i3] += B[i2-1][i3-1];
		
			for(;A[i2][i3] > mod;A[i2][i3] -= mod);
		
			if(p<<1 == N+M)
				rez += (ll)A[i2][i3];
		}
		CP(B,A);
	}	
	
	printf("%lld\n",(rez>>1) % mod);
}

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