Count EVEN and ODD numbers

I have a problem where i needed to identify 5 input number whether they are EVEN or ODD which I am able to do. The problem is I needed the Total Count of the EVEN and ODD which is the thing that I do not know.

This is the code for inputs. I just need the code for the Total Count.

a = input("Please enter 1st number: ");

if mod(a,2)==0
    disp("The 1st number is EVEN");
else
    disp("The 1st number is ODD");
end

b = input("Please enter 2nd number: ");

if mod(b,2)==0
    disp("The 2nd number is EVEN");
else
    disp("The 2nd number is ODD");
end

c = input("Please enter 3rd number: ");

if mod(c,2)==0
    disp("The 3rd number is EVEN");
else
    disp("The 3rd number is ODD");
end

d = input("Please enter 4th number: ");

if mod(d,2)==0
    disp("The 4th number is EVEN");
else
    disp("The 4th number is ODD");
end

e = input("Please enter 5th number: ");

if mod(e,2)==0
    disp("The 5th number is EVEN");
else
    disp("The 5th number is ODD");
end

Answer 1

[KSSV]

clc; clear all ;
a = input("Please enter 1st number: ");
nE = 0 ; nO = 0 ;
if mod(a,2)==0
    disp("The 1st number is EVEN");
    nE = nE+1 ;
else
    disp("The 1st number is ODD");
    nO = nO+1 ;
end

b = input("Please enter 2nd number: ");

if mod(b,2)==0
    disp("The 2nd number is EVEN");
    nE = nE+1 ;
else
    disp("The 2nd number is ODD");
    nO = nO+1 ;
end

c = input("Please enter 3rd number: ");

if mod(c,2)==0
    disp("The 3rd number is EVEN");
    nE = nE+1 ;
else
    disp("The 3rd number is ODD");
    nO = nO+1 ;
end

d = input("Please enter 4th number: ");

if mod(d,2)==0
    disp("The 4th number is EVEN");
    nE = nE+1 ;
else
    disp("The 4th number is ODD");
    nO = nO+1 ;
end

e = input("Please enter 5th number: ");

if mod(e,2)==0
    disp("The 5th number is EVEN");
    nE = nE+1 ;
else
    disp("The 5th number is ODD");
    nO = nO+1 ;
end
fprintf('there are %d number of Even Numbers\n',nE)
fprintf('there are %d number of Odd Numbers\n',nO)

[Jan Vergel De Ocampo]

Thanks a lot. Didn't even think about that.

[KSSV]

Thanks is accepting/ voting the answer :)