#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e9 + 7;
const long long sz = 1e5 + 20;
const long long N = 1e5 + 20;
void solve() {
  long long n, w = 0, b = 0, x;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    if (i & 1)
      w += (x + 1) / 2, b += (x / 2);
    else
      w += (x / 2), b += (x + 1) / 2;
  }
  cout << min(b, w) << endl;
  return;
}
int32_t main() {
  clock_t st = clock();
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
  clog << (long double)((clock() - st) / CLOCKS_PER_SEC) << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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;
const long long INF = 1e9 + 7;
const long long sz = 1e5 + 20;
const long long N = 1e5 + 20;
void solve() {
  long long n, w = 0, b = 0, x;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    if (i & 1)
      w += (x + 1) / 2, b += (x / 2);
    else
      w += (x / 2), b += (x + 1) / 2;
  }
  cout << min(b, w) << endl;
  return;
}
int32_t main() {
  clock_t st = clock();
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    solve();
  }
  clog << (long double)((clock() - st) / CLOCKS_PER_SEC) << endl;
  return 0;
}

```