#include <bits/stdc++.h>
using namespace std;
int a[300010];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  long long ans = 0;
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    ans += a[i] / 2;
    if (a[i] % 2 != 0) {
      v.push_back(i);
    }
  }
  stack<int> q;
  if (v.size() == 0) {
    cout << ans << endl;
    return 0;
  }
  q.push(v[0]);
  for (int i = 1; i < v.size(); ++i) {
    if (q.size() && (v[i] - q.top()) % 2 != 0) {
      q.pop();
      ans++;
      continue;
    }
    q.push(v[i]);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
int a[300010];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  long long ans = 0;
  vector<int> v;
  for (int i = 1; i <= n; ++i) {
    ans += a[i] / 2;
    if (a[i] % 2 != 0) {
      v.push_back(i);
    }
  }
  stack<int> q;
  if (v.size() == 0) {
    cout << ans << endl;
    return 0;
  }
  q.push(v[0]);
  for (int i = 1; i < v.size(); ++i) {
    if (q.size() && (v[i] - q.top()) % 2 != 0) {
      q.pop();
      ans++;
      continue;
    }
    q.push(v[i]);
  }
  cout << ans << endl;
  return 0;
}

```