You are given a Young diagram. 

Given diagram is a histogram with n columns of lengths a_1, a_2, …, a_n (a_1 ≥ a_2 ≥ … ≥ a_n ≥ 1).

<image> Young diagram for a=[3,2,2,2,1].

Your goal is to find the largest number of non-overlapping dominos that you can draw inside of this histogram, a domino is a 1 × 2 or 2 × 1 rectangle.

Input

The first line of input contain one integer n (1 ≤ n ≤ 300 000): the number of columns in the given histogram.

The next line of input contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 300 000, a_i ≥ a_{i+1}): the lengths of columns.

Output

Output one integer: the largest number of non-overlapping dominos that you can draw inside of the given Young diagram.

Example

Input


5
3 2 2 2 1


Output


4

Note

Some of the possible solutions for the example:

<image> <image>
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
using ll = long long;
using ull = unsigned long long;
using ld = long double;
constexpr ll INF = 2e18;
constexpr ld EPS = 1e-9;
constexpr int MOD = 1e9 + 7;
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
template <typename T1, typename T2>
using safe_map = unordered_map<T1, T2, custom_hash>;
template <typename T1, typename T2>
istream &operator>>(istream &istream, pair<T1, T2> &p) {
  istream >> p.first >> p.second;
  return istream;
}
template <typename T>
istream &operator>>(istream &istream, vector<T> &v) {
  for (auto &it : v) cin >> it;
  return istream;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &ostream, const pair<T1, T2> &p) {
  ostream << p.first << " " << p.second;
  return ostream;
}
template <typename T>
ostream &operator<<(ostream &ostream, const vector<T> &c) {
  for (auto &it : c) cout << it << " ";
  return ostream;
}
template <typename T>
void print(T &&t) {
  cout << t << "\n";
}
template <typename T, typename... Args>
void print(T &&t, Args &&...args) {
  cout << t << " ";
  print(forward<Args>(args)...);
}
template <typename T>
int32_t size_i(T &container) {
  return static_cast<int32_t>(container.size());
}
long long GCD(long long a, long long b) {
  if (!b) return a;
  return GCD(b, a % b);
}
long long LCM(long long a, long long b) { return (a * b) / GCD(a, b); }
long long modpow(ll x, long long n, long long m = MOD) {
  ll res = 1;
  while (n > 0) {
    if (n & 1) res = (res * x) % m;
    x = (x * x) % m;
    n >>= 1;
  }
  return res;
}
long long binpow(long long x, long long n) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = (x * x);
    n >>= 1;
  }
  return res;
}
long long modinv(long long x, long long m = MOD) { return modpow(x, m - 2, m); }
mt19937 rng;
long long getRandomNumber(long long l, long long r) {
  uniform_int_distribution<long long> dist(l, r);
  return dist(rng);
}
void preSolve() {
  rng = mt19937(chrono::steady_clock::now().time_since_epoch().count());
}
void solve2(long long &x, long long &y, long long ai) {
  x += (ai + 1) / 2;
  y += ai / 2;
}
void solve(long long tc) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  cin >> a;
  long long x = 0, y = 0;
  long long cnt = 0;
  for (long long i = 0; i < n; i++) {
    if (cnt == 0)
      solve2(x, y, a[i]);
    else
      solve2(y, x, a[i]);
    cnt ^= 1;
  }
  print(min(x, y));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(12) << fixed;
  preSolve();
  long long tests = 1;
  for (long long t = 1; t <= tests; t++) solve(t);
  return 0;
}
