#include <bits/stdc++.h>
using namespace std;
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const pair<Arg1, Arg2>& x) {
  return out << "(" << x.first << "," << x.second << ")";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const vector<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const set<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const map<Arg1, Arg2>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << " ] " << endl;
}
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args) {
  const string name = names.substr(0, names.find(','));
  cerr << name << " : " << arg1 << " | ";
  __trace__(names.substr(1 + (int)name.size()), args...);
}
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(12);
  cout << fixed;
  int n;
  cin >> n;
  vector<long long> a(n);
  long long ans = 0;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i & 1)
      ans += a[i] / 2;
    else
      ans += (a[i] + 1) / 2;
    sum += a[i];
  }
  ans = min(ans, sum - ans);
  cout << ans << '\n';
  time_taken();
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const pair<Arg1, Arg2>& x) {
  return out << "(" << x.first << "," << x.second << ")";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const vector<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
ostream& operator<<(ostream& out, const set<Arg1>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1, typename Arg2>
ostream& operator<<(ostream& out, const map<Arg1, Arg2>& a) {
  out << "[";
  for (const auto& x : a) out << x << ",";
  return out << "]";
}
template <typename Arg1>
void __trace__(const string name, Arg1&& arg1) {
  cerr << name << " : " << arg1 << " ] " << endl;
}
template <typename Arg1, typename... Args>
void __trace__(const string names, Arg1&& arg1, Args&&... args) {
  const string name = names.substr(0, names.find(','));
  cerr << name << " : " << arg1 << " | ";
  __trace__(names.substr(1 + (int)name.size()), args...);
}
clock_t time_p = clock();
void time_taken() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
const long long mod = 1e9 + 7;
const long long INF = 1e18;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout.precision(12);
  cout << fixed;
  int n;
  cin >> n;
  vector<long long> a(n);
  long long ans = 0;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (i & 1)
      ans += a[i] / 2;
    else
      ans += (a[i] + 1) / 2;
    sum += a[i];
  }
  ans = min(ans, sum - ans);
  cout << ans << '\n';
  time_taken();
  return 0;
}

```