#include <bits/stdc++.h>
using namespace std;
int n;
long long cnt[2] = {0, 0};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    cnt[i % 2] += x / 2;
    cnt[!(i % 2)] += x - x / 2;
  }
  cout << min(cnt[0], cnt[1]);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 n;
long long cnt[2] = {0, 0};
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    cnt[i % 2] += x / 2;
    cnt[!(i % 2)] += x - x / 2;
  }
  cout << min(cnt[0], cnt[1]);
  return 0;
}

```