Problem: To defeat Lord Voldemort, Harry needs to destroy all horcruxes first. The last horcrux is an array a of n integers, which also needs to be destroyed. The array is considered destroyed if all its elements are zeroes. To destroy the array, Harry can perform two types of operations:

  1. choose an index i (1 ≤ i ≤ n), an integer x, and subtract x from a_i.
  2. choose two indices i and j (1 ≤ i, j ≤ n; i ≠ j), an integer x, and subtract x from a_i and x + 1 from a_j. 



Note that x does not have to be positive.

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Harry is in a hurry, please help him to find the minimum number of operations required to destroy the array and exterminate Lord Voldemort.

Input

The first line contains a single integer n — the size of the array a (1 ≤ n ≤ 20). 

The following line contains n integers a_1, a_2, …, a_n — array elements (-10^{15} ≤ a_i ≤ 10^{15}).

Output

Output a single integer — the minimum number of operations required to destroy the array a.

Examples

Input


3
1 10 100


Output


3


Input


3
5 3 -2


Output


2


Input


1
0


Output


0

Note

In the first example one can just apply the operation of the first kind three times.

In the second example, one can apply the operation of the second kind two times: first, choose i = 2, j = 1, x = 4, it transforms the array into (0, -1, -2), and then choose i = 3, j = 2, x = -2 to destroy the array.

In the third example, there is nothing to be done, since the array is already destroyed.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <class t, class u>
void chmax(t& first, u second) {
  if (first < second) first = second;
}
template <class t, class u>
void chmin(t& first, u second) {
  if (second < first) first = second;
}
template <class t>
using vc = vector<t>;
template <class t>
using vvc = vc<vc<t>>;
using pi = pair<int, int>;
using vi = vc<int>;
template <class t, class u>
ostream& operator<<(ostream& os, const pair<t, u>& p) {
  return os << "{" << p.first << "," << p.second << "}";
}
template <class t>
ostream& operator<<(ostream& os, const vc<t>& v) {
  os << "{";
  for (auto e : v) os << e << ",";
  return os << "}";
}
using uint = unsigned;
using ull = unsigned long long;
template <class t, size_t n>
ostream& operator<<(ostream& os, const array<t, n>& first) {
  return os << vc<t>(first.begin(), first.end());
}
template <int i, class T>
void print_tuple(ostream&, const T&) {}
template <int i, class T, class H, class... Args>
void print_tuple(ostream& os, const T& t) {
  if (i) os << ",";
  os << get<i>(t);
  print_tuple<i + 1, T, Args...>(os, t);
}
template <class... Args>
ostream& operator<<(ostream& os, const tuple<Args...>& t) {
  os << "{";
  print_tuple<0, tuple<Args...>, Args...>(os, t);
  return os << "}";
}
template <class t>
void print(t x, int suc = 1) {
  cout << x;
  if (suc == 1) cout << "\n";
  if (suc == 2) cout << " ";
}
ll read() {
  ll i;
  cin >> i;
  return i;
}
vi readvi(int n, int off = 0) {
  vi v(n);
  for (int i = int(0); i < int(n); i++) v[i] = read() + off;
  return v;
}
template <class T>
void print(const vector<T>& v, int suc = 1) {
  for (int i = int(0); i < int(v.size()); i++)
    print(v[i], i == int(v.size()) - 1 ? suc : 2);
}
string readString() {
  string s;
  cin >> s;
  return s;
}
template <class T>
T sq(const T& t) {
  return t * t;
}
void yes(bool ex = true) {
  cout << "Yes"
       << "\n";
  if (ex) exit(0);
}
void no(bool ex = true) {
  cout << "No"
       << "\n";
  if (ex) exit(0);
}
void possible(bool ex = true) {
  cout << "Possible"
       << "\n";
  if (ex) exit(0);
}
void impossible(bool ex = true) {
  cout << "Impossible"
       << "\n";
  if (ex) exit(0);
}
constexpr ll ten(int n) { return n == 0 ? 1 : ten(n - 1) * 10; }
const ll infLL = LLONG_MAX / 3;
const int inf = INT_MAX / 2 - 100;
int topbit(signed t) { return t == 0 ? -1 : 31 - __builtin_clz(t); }
int topbit(ll t) { return t == 0 ? -1 : 63 - __builtin_clzll(t); }
int botbit(signed first) { return first == 0 ? 32 : __builtin_ctz(first); }
int botbit(ll first) { return first == 0 ? 64 : __builtin_ctzll(first); }
int popcount(signed t) { return __builtin_popcount(t); }
int popcount(ll t) { return __builtin_popcountll(t); }
bool ispow2(int i) { return i && (i & -i) == i; }
int mask(int i) { return (int(1) << i) - 1; }
bool inc(int first, int second, int c) {
  return first <= second && second <= c;
}
template <class t>
void mkuni(vc<t>& v) {
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
}
ll rand_int(ll l, ll r) {
  static random_device rd;
  static mt19937_64 gen(rd());
  return uniform_int_distribution<ll>(l, r)(gen);
}
template <class t>
int lwb(const vc<t>& v, const t& first) {
  return lower_bound(v.begin(), v.end(), first) - v.begin();
}
const int nmax = 20;
bool ok[1 << nmax], subok[1 << nmax];
ll sum[1 << nmax];
int dp[1 << nmax];
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(0);
  cout << fixed << setprecision(20);
  int n;
  cin >> n;
  vc<ll> first(n);
  for (int i = int(0); i < int(n); i++) cin >> first[i];
  for (int bit = int(1); bit < int(1 << n); bit++) {
    for (int i = int(0); i < int(n); i++)
      if (bit & 1 << i) sum[bit] += first[i];
  }
  int hasok = 0;
  for (int bit = int(1); bit < int(1 << n); bit++)
    if (!subok[bit]) {
      int sub = bit;
      sub ^= 1 << topbit(bit);
      int k = popcount(bit) - 1;
      while (sub) {
        ll s = sum[bit] - sum[sub] * 2;
        if (-k <= s && s <= k && (s & 1) == (k & 1)) {
          ok[bit] = true;
          break;
        }
        sub = (sub - 1) & bit;
      }
      if (k == 0 && sum[bit] == 0) ok[bit] = 1;
      if (ok[bit]) {
        hasok |= bit;
        for (int i = int(0); i < int(n); i++)
          if ((bit & 1 << i) == 0) subok[bit ^ 1 << i] = 1;
      }
    }
  int mx = 0;
  for (int bit = int(1); bit < int(1 << n); bit++)
    if (hasok & bit) {
      if (ok[bit]) dp[bit] = 1;
      int sub = bit;
      sub ^= 1 << topbit(bit);
      do {
        chmax(dp[bit], dp[sub] + dp[bit ^ sub]);
        sub = (sub - 1) & bit;
      } while (sub != bit);
      chmax(mx, dp[bit]);
    }
  cout << n - mx << endl;
}