#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    long long odd = 0, even = 0;
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      if (arr[i] & 1) {
        if (i & 1)
          odd++;
        else
          even++;
      }
      ans += arr[i] / 2;
    }
    cout << ans + min(odd, even) << endl;
  }
}

### Prompt

Please formulate 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    long long odd = 0, even = 0;
    long long ans = 0;
    for (long long i = 0; i < n; i++) {
      if (arr[i] & 1) {
        if (i & 1)
          odd++;
        else
          even++;
      }
      ans += arr[i] / 2;
    }
    cout << ans + min(odd, even) << endl;
  }
}

```