#include <bits/stdc++.h>
using namespace std;
const long long MaxN = 3e5 + 10;
long long a[MaxN];
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long sum1 = 0, sum2 = 0;
  for (long long i = 0; i < n; i++) {
    if (i % 2 == 0) {
      sum1 += a[i] / 2;
      sum2 += (a[i] + 1) / 2;
    } else {
      sum1 += (a[i] + 1) / 2;
      sum2 += a[i] / 2;
    }
  }
  cout << min(sum1, sum2);
}

### 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;
const long long MaxN = 3e5 + 10;
long long a[MaxN];
int main() {
  long long n;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long sum1 = 0, sum2 = 0;
  for (long long i = 0; i < n; i++) {
    if (i % 2 == 0) {
      sum1 += a[i] / 2;
      sum2 += (a[i] + 1) / 2;
    } else {
      sum1 += (a[i] + 1) / 2;
      sum2 += a[i] / 2;
    }
  }
  cout << min(sum1, sum2);
}

```