#include <bits/stdc++.h>
using namespace std;
const double PI = acos(-1);
const long long mod = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  while (test--) {
    int n;
    cin >> n;
    long long x = 0, a = 0, b = 0;
    for (int i = 0; i < n; ++i) {
      cin >> x;
      a += x / 2;
      b += x / 2;
      if (i & 1) {
        if (x & 1) a++;
      } else {
        if (x & 1) b++;
      }
    }
    cout << min(a, b) << "\n";
  }
  return 0;
}

### Prompt

Generate 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;
const double PI = acos(-1);
const long long mod = 1e9 + 7;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int test = 1;
  while (test--) {
    int n;
    cin >> n;
    long long x = 0, a = 0, b = 0;
    for (int i = 0; i < n; ++i) {
      cin >> x;
      a += x / 2;
      b += x / 2;
      if (i & 1) {
        if (x & 1) a++;
      } else {
        if (x & 1) b++;
      }
    }
    cout << min(a, b) << "\n";
  }
  return 0;
}

```