// 遅めの解法（O(T log_2(lfloor log_2 (max {2,N}) rfloor !))解）チェック
// 残念ながらc++ならこれでも通る。
#pragma GCC optimize ( "O3" )
#pragma GCC optimize( "unroll-loops" )
#pragma GCC target ( "sse4.2,fma,avx2,popcnt,lzcnt,bmi2" )
#include <bits/stdc++.h>
using namespace std;

using ll = long long;

#define MAIN main
#define TYPE_OF( VAR ) remove_const<remove_reference<decltype( VAR )>::type >::type
#define UNTIE ios_base::sync_with_stdio( false ); cin.tie( nullptr )
#define CEXPR( LL , BOUND , VALUE ) constexpr const LL BOUND = VALUE
#define CIN( LL , A ) LL A; cin >> A
#define ASSERT( A , MIN , MAX ) assert( ( MIN ) <= A && A <= ( MAX ) )
#define CIN_ASSERT( A , MIN , MAX ) CIN( TYPE_OF( MAX ) , A ); ASSERT( A , MIN , MAX )
#define FOR( VAR , INITIAL , FINAL_PLUS_ONE ) for( TYPE_OF( FINAL_PLUS_ONE ) VAR = INITIAL ; VAR < FINAL_PLUS_ONE ; VAR ++ )
#define REPEAT( HOW_MANY_TIMES ) FOR( VARIABLE_FOR_REPEAT ## HOW_MANY_TIMES , 0 , HOW_MANY_TIMES )
#define QUIT return 0
#define COUT( ANSWER ) cout << ( ANSWER ) << "\n"
#define RETURN( ANSWER ) COUT( ANSWER ); QUIT

int g( ll n )
{
  bool found[60] = {};
  int d = 0;
  while( n != 0 ){
    if( ( n & 1 ) == 1 ){
      found[ g( d ) ] = true;
    }
    d++;
    n >>= 1;
  }
  int gn = 0;
  while( found[gn] ){
    gn++;
  }
  return gn;
}

int MAIN()
{
  UNTIE;
  CEXPR( int , bound_N , 100000 );
  CIN_ASSERT( N , 1 , bound_N );
  CEXPR( ll , bound_Ai , 1000000000000000000 );
  bool even = true;
  int grundy = 0;
  REPEAT( N ){
    CIN_ASSERT( Ai , -bound_Ai , bound_Ai );
    if( Ai < 0 ){
      even = !even;
    } else {
      grundy ^= g( Ai );
    }
  }
  RETURN( ( even && grundy == 0 ) ? 2 : 1 );
}