/* Solution to Topcoder SRM 699, D1, 250-pointer */
/* James S. Plank */
/* October 6, 2016 */

#include <vector>
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;

class OthersXor {
  public:
    long long minSum(vector <int> x);
};

long long OthersXor::minSum(vector <int> x)
{
  bool U;
  int T;
  vector <int> W;           // These are the integers W
  vector <int> Wbit;        // These will be the bits of row i of W.
  vector <int> A;
  int R;
  int B;
  int Best_B;               // We keep track of the best B for each row.
  int i, j;
  long long two_to_the_i;
  long long RV;

  RV = 0;
  two_to_the_i = 1;

  /* First, calculate U and the W's */

  U = false;
  for (i = 0; i < x.size(); i++) {
    if (x[i] == -1) {
      U = true;
    } else {
      W.push_back(x[i]);
    }
  }

  /* Next, concentrate on each bit individually */

  for (i = 0; i < 32; i++) {

    /* Grab the bits of W */

    Wbit.clear();
    for (j = 0; j < W.size(); j++) {
      if (W[j] & (1 << i)) {
        Wbit.push_back(1);
      } else {
        Wbit.push_back(0);
      }
    }

    /* Sentinelize Best_B */

    Best_B = 500;
    
    /* Do as in the writeup.  First calculate A, then R, then B, for each setting of T */

    for (T = 0; T < 2; T++) {

      printf("W:");
      for (j = 0; j < Wbit.size(); j++) printf(" %d", Wbit[j]);

      /* Here, we calculate A. */

      A.clear();
      for (j = 0; j < Wbit.size(); j++) A.push_back(T ^ Wbit[j]);
      printf("   A:");
      for (j = 0; j < Wbit.size(); j++) printf(" %d", A[j]);

      /* Now R and B */

      R = T;
      B = 0;
      for (j = 0; j < Wbit.size(); j++) {
        R ^= A[j];
        B += A[j];
      }
      B += R;

      /* Check to see if B is legal, and if so check it against Best_B */

      if (R == 1 && U == false) {
        /* The answer is illegal */
      } else if (B < Best_B) {
        Best_B = B;
      }
      printf("   R:%d   i:%d T:%d B:%d\n", R, i, T, B);
    }

    /* If none of the B's were legal, then return -1 */

    if (Best_B == 500) return -1;

    /* Otherwise, add B times 2^i to RV */

    RV += (two_to_the_i * Best_B);
    two_to_the_i <<= 1;
  }

  return RV;
}