my output keeps getting $"$No optimal solution" but it has one and I need it to solve my sudoku puzzle here's my code to help fix:

import pyomo.environ as pyo

# Define the model and grid parameters

model $=$ pyo.ConcreteModel$()$

N $=9$ # Grid size for standard Sudoku

values$\_$range $=$ range$(1,$ N $+1)$

rows$\_$range $=$ range$($N$)$

cols$\_$range $=$ range$($N$)$

# Decision variable: x$[$r$,$ c$,$ n$]$ is $1$ if number n is in cell $($r$,$ c$),0$ otherwise

model.x $=$ pyo.Var$($rows$\_$range, cols$\_$range, values$\_$range, domain$=$pyo.Binary$)$

model.constraints $=$ pyo.ConstraintList$()$

# Constraint $1$: Each cell contains one number

for r in rows$\_$range:

for c in cols$\_$range:

model.constraints.add$($sum$($model$.$x$[$r$,$ c$,$ n$]$ for n in values$\_$range$)==1)$

# Constraint $2$: Each row contains unique values

for r in rows$\_$range:

for n in values$\_$range:

model.constraints.add$($sum$($model$.$x$[$r$,$ c$,$ n$]$ for c in cols$\_$range$)==1)$

# Constraint $3$: Each column contains unique values

for c in cols$\_$range:

for n in values$\_$range:

model.constraints.add$($sum$($model$.$x$[$r$,$ c$,$ n$]$ for r in rows$\_$range$)==1)$

# Constraint $4$: Each $3$x$3$ block contains unique values

sqrtN $=3$

for br in range$(0,$ N$,$ sqrtN$)$:

for bc in range$(0,$ N$,$ sqrtN$)$:

for n in values$\_$range:

model.constraints.add$($

sum$($model$.$x$[$r$,$ c$,$ n$]$ for r in range$($br$,$ br $+$ sqrtN$)$ for c in range$($bc$,$ bc $+$ sqrtN$))==1$

$)$

# Constraint $5$: Pre$-$filled values

prefilled$\_$cells $=\{$

$(0,0,8),$

$(1,2,3),(1,3,6),$

$(2,1,7),(2,4,9),(2,6,2),$

$(3,0,5),(3,5,7),$

$(4,4,4),(4,5,5),(4,6,7),$

$(5,3,1),(5,7,3),$

$(6,3,1),(6,7,6),(6,8,8),$

$(7,2,8),(7,3,5),(7,7,1),$

$(8,1,9),(8,6,4)$

$\}$

for $($r$,$ c$,$ n$)$ in prefilled$\_$cells:

model.constraints.add$($model$.$x$[$r$,$ c$,$ n$]==1)$

# Solve the model

result $=$ pyo.SolverFactory$("$glpk$").$solve$($model$)$

# Print results

if result.solver.termination$\_$condition $==$ pyo.TerminationCondition.optimal:

print$("$Solved Sudoku Puzzle:"$)$

for r in rows$\_$range:

row $=[]$

for c in cols$\_$range:

for n in values$\_$range:

if model.x$[$r$,$ c$,$ n$]==1$:

row.append$($n$)$

break

print$("".$join$($str$($num$)$ for num in row$))$

else:

print$("$No optimal solution found."$)$