#include <bits/stdc++.h>
using namespace std;
void Weapons19() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  long long int ans = 0, p = 0;
  stack<long long int> s;
  for (long long int i = 0; i < n; i++) {
    if (v[i] % 2 == 0) {
      if (!s.empty() && s.top() % 2 == 0) {
        long long int k = s.top();
        s.pop();
        ans += (v[i] + k) / 2;
      } else
        s.push(v[i]);
    } else {
      if (!s.empty() && s.top() % 2 == 1) {
        long long int k = s.top();
        s.pop();
        ans += (v[i] + k) / 2;
      } else
        s.push(v[i]);
    }
  }
  long long int k = 0;
  while (!s.empty()) {
    ans += s.top() / 2;
    s.pop();
  }
  cout << ans;
}
int32_t main() {
  cout << fixed << setprecision(16);
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Weapons19();
}

### 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;
void Weapons19() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  for (long long int i = 0; i < n; i++) cin >> v[i];
  long long int ans = 0, p = 0;
  stack<long long int> s;
  for (long long int i = 0; i < n; i++) {
    if (v[i] % 2 == 0) {
      if (!s.empty() && s.top() % 2 == 0) {
        long long int k = s.top();
        s.pop();
        ans += (v[i] + k) / 2;
      } else
        s.push(v[i]);
    } else {
      if (!s.empty() && s.top() % 2 == 1) {
        long long int k = s.top();
        s.pop();
        ans += (v[i] + k) / 2;
      } else
        s.push(v[i]);
    }
  }
  long long int k = 0;
  while (!s.empty()) {
    ans += s.top() / 2;
    s.pop();
  }
  cout << ans;
}
int32_t main() {
  cout << fixed << setprecision(16);
  cin.sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  Weapons19();
}

```