Explicit Finite Difference method to price European Call in Python

Question

import numpy as np
import matplotlib.pyplot as plt

def explicit_fd(isCall, S0, K, r, T, sigma, Smax, M, N):
    dt = T/N
    ds = Smax/M
    iValues = np.arange(M)
    jValues = np.arange(N)
    S = iValues*ds
    t = jValues*dt
    
    alpha = 0.5*sigma**2*dt/ds**2
    beta = 0.5*r*dt/ds
    gamma = r*dt

    # Initialize the grid
    V = np.zeros(shape=(M+1, N+1))
    S = np.zeros(M+1)

    # Terminal condition at time T
    for i in range(0, M+1):
        S[i] = i*ds
        V[i,N] = max(S[i]-K, 0) if isCall else max(K-S[i], 0)

    # Explicit finite difference method
    for j in reversed(range(0, N)):
        for i in range(1, M):
            V[i,j] = alpha*V[i+1,j+1] + (1-2*alpha-beta+gamma)*V[i,j+1] + alpha*V[i-1,j+1] + beta*V[i+1,j+1]
        # Apply boundary conditions
        V[0,j] = 0
        V[M,j] = Smax - K*np.exp(-r*(N-j)*dt) if isCall else 0

    return np.interp(S0, S, V[:,0])

T = 1 
S0 = 34.523 
K = 32.9352
sigma = 0.32 
r = 0.094  
S_max = K*2
N = 2000
M = 2000

call_price = explicit_fd(True, S0, K, r, T, sigma, S_max, M, N)
print(f"European Call Option Price: {call_price}")

Im trying to price an European Call with the Explicit Finite Difference method but the price I get is very different to the one calculated with Black-Scholes.

Price with Black Scholes = 6.757338059973058, Precio Crank Nicolson = 6.691067076466618

When I calculate it with the code above it gives me a price of 1.7959190949954134. I dont know where the error is.