#include <bits/stdc++.h>
using namespace std;
long long solve() {
  long long n, a, s = 0, o = 0, e = 0;
  vector<long long> v, u;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a;
    s = s + a / 2;
    if (a % 2) {
      v.push_back(i);
    }
  }
  for (long long i = 1; i < v.size(); i++) {
    if ((v[i] - v[i - 1]) % 2 == 1) {
      s++;
      i++;
      if (i == v.size() - 1) {
        if (v[i] % 2 == 1)
          o++;
        else
          e++;
      }
    } else if (i == v.size() - 1) {
      if (v[i] % 2 == 1)
        o++;
      else
        e++;
      if (v[i - 1] % 2 == 1)
        o++;
      else
        e++;
    } else {
      if (v[i - 1] % 2 == 1)
        o++;
      else
        e++;
    }
  }
  cout << s + min(o, e);
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

### Prompt

In CPP, your task is to solve 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;
long long solve() {
  long long n, a, s = 0, o = 0, e = 0;
  vector<long long> v, u;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a;
    s = s + a / 2;
    if (a % 2) {
      v.push_back(i);
    }
  }
  for (long long i = 1; i < v.size(); i++) {
    if ((v[i] - v[i - 1]) % 2 == 1) {
      s++;
      i++;
      if (i == v.size() - 1) {
        if (v[i] % 2 == 1)
          o++;
        else
          e++;
      }
    } else if (i == v.size() - 1) {
      if (v[i] % 2 == 1)
        o++;
      else
        e++;
      if (v[i - 1] % 2 == 1)
        o++;
      else
        e++;
    } else {
      if (v[i - 1] % 2 == 1)
        o++;
      else
        e++;
    }
  }
  cout << s + min(o, e);
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}

```