#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
int main() {
  int n;
  ll ans = 0, bal = 0;
  scanf("%d", &n);
  vi a(n), bad;
  for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
  for (int i = 0; i < (int)(n); i++) {
    if (a[i] & 1) bad.push_back(i);
    ans += a[i];
    if (a[i] & 1) bal += (i & 1) ? 1 : -1;
  }
  printf("%lld\n", (ans - abs(bal)) / 2);
  return 0;
}

### Prompt

Your task is to create 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;
using ll = long long;
using vi = vector<int>;
using pii = pair<int, int>;
int main() {
  int n;
  ll ans = 0, bal = 0;
  scanf("%d", &n);
  vi a(n), bad;
  for (int i = 0; i < (int)(n); i++) scanf("%d", &a[i]);
  for (int i = 0; i < (int)(n); i++) {
    if (a[i] & 1) bad.push_back(i);
    ans += a[i];
    if (a[i] & 1) bal += (i & 1) ? 1 : -1;
  }
  printf("%lld\n", (ans - abs(bal)) / 2);
  return 0;
}

```