#include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
void print_time() {}
template <typename T>
void update_max(T& x, T y) {
  x = max(x, y);
}
template <typename T>
void update_min(T& x, T y) {
  x = min(x, y);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
int main() {
  cout << fixed << setprecision(20);
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  long long suma = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    suma += a[i];
  }
  long long black = 0;
  for (long long i = 0; i < n; ++i) {
    black += (a[i] + (i & 1)) / 2;
  }
  cout << min(black, suma - black);
  print_time();
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
void print_time() {}
template <typename T>
void update_max(T& x, T y) {
  x = max(x, y);
}
template <typename T>
void update_min(T& x, T y) {
  x = min(x, y);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> pair) {
  return out << "(" << pair.first << ", " << pair.second << ")";
}
template <class T>
ostream& operator<<(ostream& out, vector<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class T>
ostream& operator<<(ostream& out, set<T> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
template <class L, class R>
ostream& operator<<(ostream& out, map<L, R> vec) {
  out << "(";
  for (auto& v : vec) out << v << ", ";
  return out << ")";
}
int main() {
  cout << fixed << setprecision(20);
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  long long suma = 0;
  for (long long i = 0; i < n; ++i) {
    cin >> a[i];
    suma += a[i];
  }
  long long black = 0;
  for (long long i = 0; i < n; ++i) {
    black += (a[i] + (i & 1)) / 2;
  }
  cout << min(black, suma - black);
  print_time();
  return 0;
}

```