#include <bits/stdc++.h>
using namespace std;
const int N = 3e5;
int a[N + 100];
long long cnt[2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; i++) {
    if (i % 2) {
      cnt[1] += a[i] / 2;
      cnt[0] += (a[i] + 1) / 2;
    } else {
      cnt[0] += a[i] / 2;
      cnt[1] += (a[i] + 1) / 2;
    }
  }
  printf("%lld\n", min(cnt[0], cnt[1]));
}

### Prompt

Create a solution in CPP for 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;
const int N = 3e5;
int a[N + 100];
long long cnt[2];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
  }
  for (int i = 1; i <= n; i++) {
    if (i % 2) {
      cnt[1] += a[i] / 2;
      cnt[0] += (a[i] + 1) / 2;
    } else {
      cnt[0] += a[i] / 2;
      cnt[1] += (a[i] + 1) / 2;
    }
  }
  printf("%lld\n", min(cnt[0], cnt[1]));
}

```