#include <bits/stdc++.h>
using namespace std;
int arr[100010];
int main() {
  long long a, b, c, d, e, f, g, h, i, j, k, l = 0, m, n, o, p, q, r = 0, s, t,
                                             u, v, w, x, y, z, tmp, ans, cnt;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &tmp);
    if (tmp % 2 == 0) {
      l += tmp / 2;
      r += tmp / 2;
    }
    if (tmp % 2 == 1 && i % 2 == 0) {
      l += tmp / 2 + 1;
      r += tmp / 2;
    }
    if (tmp % 2 == 1 && i % 2 == 1) {
      l += tmp / 2;
      r += tmp / 2 + 1;
    }
  }
  printf("%lld", (l < r ? l : r));
}

### Prompt

Create a solution in CPP for 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;
int arr[100010];
int main() {
  long long a, b, c, d, e, f, g, h, i, j, k, l = 0, m, n, o, p, q, r = 0, s, t,
                                             u, v, w, x, y, z, tmp, ans, cnt;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &tmp);
    if (tmp % 2 == 0) {
      l += tmp / 2;
      r += tmp / 2;
    }
    if (tmp % 2 == 1 && i % 2 == 0) {
      l += tmp / 2 + 1;
      r += tmp / 2;
    }
    if (tmp % 2 == 1 && i % 2 == 1) {
      l += tmp / 2;
      r += tmp / 2 + 1;
    }
  }
  printf("%lld", (l < r ? l : r));
}

```