A flower shop has got n bouquets, and the i-th bouquet consists of ai flowers. Vasya, the manager of the shop, decided to make large bouquets from these bouquets. 

Vasya thinks that a bouquet is large if it is made of two or more initial bouquets, and there is a constraint: the total number of flowers in a large bouquet should be odd. Each of the initial bouquets can be a part of at most one large bouquet. If an initial bouquet becomes a part of a large bouquet, all its flowers are included in the large bouquet.

Determine the maximum possible number of large bouquets Vasya can make. 

Input

The first line contains a single positive integer n (1 ≤ n ≤ 105) — the number of initial bouquets.

The second line contains a sequence of integers a1, a2, ..., an (1 ≤ ai ≤ 106) — the number of flowers in each of the initial bouquets.

Output

Print the maximum number of large bouquets Vasya can make. 

Examples

Input

5
2 3 4 2 7


Output

2


Input

6
2 2 6 8 6 12


Output

0


Input

3
11 4 10


Output

1

Note

In the first example Vasya can make 2 large bouquets. For example, the first bouquet can contain the first and the fifth initial bouquets (the total number of flowers is then equal to 9), and the second bouquet can consist of the second and the third initial bouquets (the total number of flowers is then equal to 7). The fourth initial bouquet is unused in this scheme. 

In the second example it is not possible to form a single bouquet with odd number of flowers.

In the third example Vasya can make one large bouquet. For example, he can make it using all three initial bouquets. The size of the large bouquet is then equal to 11 + 4 + 10 = 25.
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const long long MAX = 1e18 + 1;
const double pi = 3.1415926535897932384626433832;
const long long M = 1e9 + 7;
long long modpow(long long x, long long n, long long mod1) {
  if (n == 0) return 1 % mod1;
  long long u = modpow(x, n / 2, mod1);
  u = (u * u) % mod1;
  if (n % 2 == 1) u = (u * x) % mod1;
  return u;
}
long long POW(long long x, long long y) { return modpow(x, y, LONG_LONG_MAX); }
long long bis(long long a[], long long l, long long r, long long x) {
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (a[m] == x) return m;
    if (a[m] < x)
      l = m + 1;
    else
      r = m - 1;
  }
  return -1;
}
long long gcd(long long x, long long y) {
  if (x == 0)
    return y;
  else
    return gcd(y % x, x);
}
long long lcm(long long x, long long y) { return (x * y) / gcd(x, y); }
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6) {
    if (n % i == 0 || n % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
bool pdrome(string s) {
  long long n = s.length();
  for (long long i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) {
      return false;
    }
  }
  return true;
}
long long fac[10] = {1};
void fact() {
  for (long long i = 1; i < 10; i++) {
    fac[i] = fac[i - 1] * i;
  }
}
bool sumdigits(long long a) {
  long long result = 0;
  while (a > 0) {
    result += a % 10;
    a /= 10;
  }
  return result;
}
long long digits(long long n) { return floor(log10(n) + 1); }
const long long Node = 1e5 + 7;
vector<long long> adj[Node];
bool vis[Node] = {false};
void addedge(long long u, long long v) {
  adj[u].push_back(v);
  adj[v].push_back(u);
}
void bfs(long long u) {
  vis[u] = true;
  list<long long> q;
  q.push_back(u);
  while (!q.empty()) {
    q.pop_front();
    for (auto x : adj[u]) {
      if (!vis[x]) {
        vis[x] = true;
        q.push_back(x);
      }
    }
  }
}
void dfs(long long u, vector<pair<long long, long long>>& v) {
  vis[u] = true;
  for (auto x : adj[u]) {
    if (!vis[x]) {
      v.push_back({u, x});
      dfs(x, v);
    }
  }
}
long long con1(string s) {
  long long res = 0;
  stringstream geek(s);
  geek >> res;
  return res;
}
string con2(long long num) {
  string stri = to_string(num);
  return stri;
}
long long countbits(long long k) {
  long long cnt = log2(k);
  return cnt;
}
long long ncr(long long n, long long r) {
  if (r > n || r < 0) return 0;
  long long ans = 1, cnt = 1;
  for (long long x = n; x > r; x--) {
    ans = ((ans % M) * (x % M)) % M;
    ans = (ans / cnt) % M;
    ans = ans % M;
    cnt++;
  }
  return ans % M;
}
long long Pnc(int n, int k) {
  long long P = 1;
  for (int i = 0; i < k; i++) P *= (n - i) % M;
  return P % M;
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
char chk(char c) {
  if (c == '1') return '0';
  return '1';
}
void solve() {
  long long n;
  cin >> n;
  long long a;
  long long cnt0 = 0, cnt1 = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a;
    if (a & 1) {
      cnt0++;
    } else
      cnt1++;
  }
  if (cnt1 >= cnt0) {
    cout << cnt0 << endl;
  } else {
    cout << cnt1 + (long long)((cnt0 - cnt1) / 3) << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
