#include <bits/stdc++.h>
const int maxn = 300000 + 10;
int a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  long long black = 0, white = 0;
  for (int i = 0; i < n; i++) {
    if (i & 1) {
      int b = a[i] / 2;
      int w = a[i] - b;
      black += b;
      white += w;
    } else {
      int w = a[i] / 2;
      int b = a[i] - w;
      black += b;
      white += w;
    }
  }
  printf("%lld\n", black < white ? black : white);
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
const int maxn = 300000 + 10;
int a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  long long black = 0, white = 0;
  for (int i = 0; i < n; i++) {
    if (i & 1) {
      int b = a[i] / 2;
      int w = a[i] - b;
      black += b;
      white += w;
    } else {
      int w = a[i] / 2;
      int b = a[i] - w;
      black += b;
      white += w;
    }
  }
  printf("%lld\n", black < white ? black : white);
  return 0;
}

```