int main() {
    int odd = 0, even = 0;
    int n; cin >> n; int a[n];
    for(int i = 0; i < n; i++) { 
        cin >> a[i];
        if(a[i]%2 == 0) even++;
        else odd++;
    }
    if(n == 1 & odd == 1) { cout << -1; return 0; }
    cout << min(odd%2, odd%2 + even%2);
    return 0;
}

def smallestSizeSubsequence(n, A):
    total = sum(A)
    result = 0 if total%2 == 0 else 1 if len(A)>1 else -1
    return result

if(total sum = even number) then print 0
else
  total sum is an odd number. 
  We know that sum of two odd numbers is always even num
  Since the total sum is odd, the number of odd numbers is   
  definetely odd, deleting one of the odd numbers would
  result in an even sum.
  If we have only one number which is odd, then print -1
  
  Simple Number Theory

Scanner in = new Scanner(System.in);
int n = in.nextInt();
int sum=0;
for(int i = 0; i < n; i++){
    sum += in.nextInt();
}
System.out.println((sum%2==0)?0:(n==1)?-1:1);