#include <bits/stdc++.h>
using namespace std;
int INF = 1e9 + 7;
long long INFLL = 1e17;
const int maxn = 3e5 + 5;
int n;
vector<int> A;
int main() {
  cin >> n;
  A = vector<int>(n);
  long long odd = 0, even = 0;
  for (int i = 0; i < (n); ++i) {
    cin >> A[i];
    odd += A[i] / 2;
    even += A[i] / 2;
    if (A[i] % 2) {
      if (i % 2) {
        odd++;
      } else {
        even++;
      }
    }
  }
  printf("%lld\n", min(odd, even));
}

### Prompt

Generate 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 INF = 1e9 + 7;
long long INFLL = 1e17;
const int maxn = 3e5 + 5;
int n;
vector<int> A;
int main() {
  cin >> n;
  A = vector<int>(n);
  long long odd = 0, even = 0;
  for (int i = 0; i < (n); ++i) {
    cin >> A[i];
    odd += A[i] / 2;
    even += A[i] / 2;
    if (A[i] % 2) {
      if (i % 2) {
        odd++;
      } else {
        even++;
      }
    }
  }
  printf("%lld\n", min(odd, even));
}

```