#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &pii);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, set<T> &v);
const int N = 2e5 + 5;
void Solve(int cas) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  long long cnt[2] = {0};
  long long now = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    cnt[now] += a[i] / 2;
    cnt[1 - now] += a[i] / 2;
    if (a[i] % 2) cnt[now]++;
    now = now ^ 1;
  }
  long long ans = min(cnt[0], cnt[1]);
  cout << ans << endl;
}
int main() {
  int t = 1, cas = 0;
  while (t--) {
    Solve(++cas);
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &pii) {
  os << "" << pii.first << " " << pii.second << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}

### 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 <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &pii);
template <class T>
ostream &operator<<(ostream &os, vector<T> &v);
template <class T>
ostream &operator<<(ostream &os, set<T> &v);
const int N = 2e5 + 5;
void Solve(int cas) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  long long cnt[2] = {0};
  long long now = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    cnt[now] += a[i] / 2;
    cnt[1 - now] += a[i] / 2;
    if (a[i] % 2) cnt[now]++;
    now = now ^ 1;
  }
  long long ans = min(cnt[0], cnt[1]);
  cout << ans << endl;
}
int main() {
  int t = 1, cas = 0;
  while (t--) {
    Solve(++cas);
  }
  return 0;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, pair<T1, T2> &pii) {
  os << "" << pii.first << " " << pii.second << " ";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, vector<T> &v) {
  os << "[ ";
  for (int i = 0; i < v.size(); i++) {
    os << v[i] << " ";
  }
  os << " ]";
  return os;
}
template <class T>
ostream &operator<<(ostream &os, set<T> &v) {
  os << "[ ";
  for (T i : v) {
    os << i << " ";
  }
  os << " ]";
  return os;
}

```