missing number

arr = [1, 2, 3, 4, 6, 7]
n = len(arr)
 
xor1 = xor2 = 0
 
for i in range(1, n+1):
    xor1 ^= i
 
for i in range(n-1):
    xor2 ^= arr[i]
 
print(xor1 ^ xor2)
 

#include <iostream>
 
using namespace std;
 
int main() {
  int arr[] = {1, 2, 3, 4, 6, 7};
  int n = sizeof(arr) / sizeof(arr[0]);
  int xor1 = 0, xor2 = 0;
 
  for (int i = 1; i <= n; i++) {
    xor1 ^= i;
  }
 
  for (int i = 0; i < n - 1; i++) {
    xor2 ^= arr[i];
  }
 
  cout << (xor1 ^ xor2);
}