No of Ways of Boolean Paranthesis DP

class Solution:
    def countWays(self, N, T):
        n=N//2
        A=[[-1]*(N+1) for i in range(N+1)]
        B=[[-1]*(N+1) for i in range(N+1)]
        def find(s,e):
            if e==s:
                return 1,0 if T[e] else 0,1
            if A[s][e]!=-1:
                return A[s][e],B[s][e]
            trues,falses=0,0
            for i in range(s+1,e+1,2):
                t1,f1=find(s,i-1)
                t2,f2=find(i+1,e)
                if T[i]=='&':
                    trues+=t1*t2
                    falses+=(f1*t2+t1*f2+f1*f2)
                elif T[i]=='|':
                    trues+=(f1*t2+t1*f2+t1*t2)
                    falses+=(f1*f2)
                else:
                    falses+=(t1*t2+f1*f2)
                    trues+=(f1*t2+t1*f2)
            A[s][e],B[s][e]=trues,falses
            return trues,falses
        return find(0,len(S)-1)[0]%1003