#include <bits/stdc++.h>
using namespace std;
int op[300005], n;
long long ans, s;
vector<int> vec;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", op + i), s += op[i];
  for (int i = 1; i <= n; i++) {
    if (i & 1)
      ans += (op[i] + 1) >> 1;
    else
      ans += op[i] - ((op[i] + 1) >> 1);
  }
  printf("%lld", min(ans, s - ans));
  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 op[300005], n;
long long ans, s;
vector<int> vec;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", op + i), s += op[i];
  for (int i = 1; i <= n; i++) {
    if (i & 1)
      ans += (op[i] + 1) >> 1;
    else
      ans += op[i] - ((op[i] + 1) >> 1);
  }
  printf("%lld", min(ans, s - ans));
  return 0;
}

```