#include <bits/stdc++.h>
using namespace std;
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> a[i];
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 0; i < n; i += 2) {
    if (a[i] % 2 == 1) {
      cnt1 += ((a[i] / 2) + 1);
      cnt2 += (a[i] / 2);
    } else {
      cnt1 += ((a[i] / 2));
      cnt2 += (a[i] / 2);
    }
  }
  for (long long i = 1; i < n; i += 2) {
    if (a[i] % 2 == 1) {
      cnt2 += ((a[i] / 2) + 1);
      cnt1 += (a[i] / 2);
    } else {
      cnt2 += ((a[i] / 2));
      cnt1 += (a[i] / 2);
    }
  }
  long long ans = min(cnt1, cnt2);
  cout << ans << '\n';
}

### Prompt

Your challenge is to write 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;
signed main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (__typeof(n) i = (0) - ((0) > (n)); i != (n) - ((0) > (n));
       i += 1 - 2 * ((0) > (n)))
    cin >> a[i];
  long long cnt1 = 0, cnt2 = 0;
  for (long long i = 0; i < n; i += 2) {
    if (a[i] % 2 == 1) {
      cnt1 += ((a[i] / 2) + 1);
      cnt2 += (a[i] / 2);
    } else {
      cnt1 += ((a[i] / 2));
      cnt2 += (a[i] / 2);
    }
  }
  for (long long i = 1; i < n; i += 2) {
    if (a[i] % 2 == 1) {
      cnt2 += ((a[i] / 2) + 1);
      cnt1 += (a[i] / 2);
    } else {
      cnt2 += ((a[i] / 2));
      cnt1 += (a[i] / 2);
    }
  }
  long long ans = min(cnt1, cnt2);
  cout << ans << '\n';
}

```