#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1e3;
long long n, a, b, w, res;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    if (a == 0) {
      res += min(b, w);
      b = 0;
      w = 0;
    } else if (a % 2 == 0) {
      b += a / 2;
      w += a / 2;
    } else {
      if (i % 2 == 1) {
        b += a / 2 + 1;
        w += a / 2;
      } else {
        b += a / 2;
        w += a / 2 + 1;
      }
    }
  }
  res += min(b, w);
  cout << res << endl;
}

### Prompt

Your challenge is to write 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 int N = 1e6 + 1e3;
long long n, a, b, w, res;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    if (a == 0) {
      res += min(b, w);
      b = 0;
      w = 0;
    } else if (a % 2 == 0) {
      b += a / 2;
      w += a / 2;
    } else {
      if (i % 2 == 1) {
        b += a / 2 + 1;
        w += a / 2;
      } else {
        b += a / 2;
        w += a / 2 + 1;
      }
    }
  }
  res += min(b, w);
  cout << res << endl;
}

```