#include <bits/stdc++.h>
int a[300050], c[2];
int main() {
  int n, i;
  long long ans = 0ll;
  scanf("%d", &n);
  for (i = 1; i <= (n); ++i) scanf("%d", a + i);
  for (i = 1; i <= (n); ++i) ans += a[i] >> 1;
  for (i = 1; i <= (n); ++i)
    if (a[i] & 1)
      if (c[(i & 1) ^ 1]) {
        ++ans;
        --c[(i & 1) ^ 1];
      } else
        ++c[i & 1];
  printf("%I64d", ans);
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
int a[300050], c[2];
int main() {
  int n, i;
  long long ans = 0ll;
  scanf("%d", &n);
  for (i = 1; i <= (n); ++i) scanf("%d", a + i);
  for (i = 1; i <= (n); ++i) ans += a[i] >> 1;
  for (i = 1; i <= (n); ++i)
    if (a[i] & 1)
      if (c[(i & 1) ^ 1]) {
        ++ans;
        --c[(i & 1) ^ 1];
      } else
        ++c[i & 1];
  printf("%I64d", ans);
  return 0;
}

```