#include <bits/stdc++.h>
using namespace std;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int INF = 1e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a[n + 1];
  long long res = 0;
  for (int i = 1; i <= n; i++) cin >> a[i], res += a[i] / 2, a[i] %= 2;
  int c = 0, l = 0;
  for (int i = 1; i <= n; i++) {
    if (!a[i]) continue;
    if (i & 1)
      l++;
    else
      c++;
  }
  cout << res + min(c, l);
  return 0;
}

### Prompt

Please formulate a CPP solution to 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;
template <typename T, typename U>
inline void amin(T &x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
inline void amax(T &x, U y) {
  if (x < y) x = y;
}
const int INF = 1e9;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  long long a[n + 1];
  long long res = 0;
  for (int i = 1; i <= n; i++) cin >> a[i], res += a[i] / 2, a[i] %= 2;
  int c = 0, l = 0;
  for (int i = 1; i <= n; i++) {
    if (!a[i]) continue;
    if (i & 1)
      l++;
    else
      c++;
  }
  cout << res + min(c, l);
  return 0;
}

```