#include <bits/stdc++.h>
using namespace std;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int n, first;
  long long c0 = 0, c1 = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> first;
    c0 += first / 2;
    c1 += first / 2;
    if (first & 1 and i & 1) c0++;
    if (first & 1 and !(i & 1)) c1++;
  }
  cout << min(c0, c1) << endl;
  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>
using namespace std;
int power(int a, int b, int m, int ans = 1) {
  for (; b; b >>= 1, a = 1LL * a * a % m)
    if (b & 1) ans = 1LL * ans * a % m;
  return ans;
}
int main() {
  int n, first;
  long long c0 = 0, c1 = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> first;
    c0 += first / 2;
    c1 += first / 2;
    if (first & 1 and i & 1) c0++;
    if (first & 1 and !(i & 1)) c1++;
  }
  cout << min(c0, c1) << endl;
  return 0;
}

```