#include <bits/stdc++.h>
long long int a[300000];
int main() {
  long long int n, i, hei, bai;
  while (scanf("%lld", &n) != EOF) {
    hei = bai = 0;
    for (i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      if (i % 2) {
        hei += (a[i] + 1) / 2;
        bai += a[i] / 2;
      } else {
        bai += (a[i] + 1) / 2;
        hei += a[i] / 2;
      }
    }
    printf("%lld\n", bai < hei ? bai : hei);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
long long int a[300000];
int main() {
  long long int n, i, hei, bai;
  while (scanf("%lld", &n) != EOF) {
    hei = bai = 0;
    for (i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
      if (i % 2) {
        hei += (a[i] + 1) / 2;
        bai += a[i] / 2;
      } else {
        bai += (a[i] + 1) / 2;
        hei += a[i] / 2;
      }
    }
    printf("%lld\n", bai < hei ? bai : hei);
  }
  return 0;
}

```