Another Palindrome

#include <bits/stdc++.h>
using namespace std;
#define PB push_back
#define ZERO (1e-10)
#define INF (1<<29)
#define CL(A,I) (memset(A,I,sizeof(A)))
#define DEB printf("DEB!\n");
#define D(X) cout<<"  "<<#X": "<<X<<endl;
#define EQ(A,B) (A+ZERO>B&&A-ZERO<B)
typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
#define IN(n) int n;scanf("%d",&n);
#define FOR(i, m, n) for (int i(m); i < n; i++)
#define REP(i, n) FOR(i, 0, n)
#define F(n) REP(i, n)
#define FF(n) REP(j, n)
#define FT(m, n) FOR(k, m, n)
#define aa first
#define bb second
void ga(int N,int *A){F(N)scanf("%d",A+i);}
#define MX (62)
char s[MX];
ll dp[MX][MX][2];
void mb(int&b,int&e,char c){
    while(s[b]^c)++b;
    while(s[e]^c)--e;
}
ll dyn(int b,int e,bool x){
    if(b==e)return x+1;
    if(b>e)return x;
    ll &v(dp[b][e][x]);
    int B=e,E=b;
    if(~v)return v;
    v=s[b]==s[e]?dyn(b+1,e-1,1):0;
    do{
        v+=dyn(b+1,E-1,1);
        while(s[++E]^s[b]&&E<e);
    }while(E<e);
    do{
        v+=dyn(B+1,e-1,1);
        while(s[e]^s[--B]&&B>b);
    }while(B>b);
    v+=dyn(b+1,e-1,x);
    return v;
}
int main(void){
    IN(tt)F(tt)
        scanf("%s",s),CL(dp,-1),printf("%lld\n",dyn(0,strlen(s)-1,0));
    return 0;
}