#include <bits/stdc++.h>
using namespace std;
long long int insig;
void Out(vector<long long int> x) {
  for (long long int i = 0; i < x.size(); i++) {
    cout << x[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int N;
  cin >> N;
  vector<long long int> a;
  for (long long int IBRO = 0; IBRO < N; IBRO++) {
    cin >> insig;
    a.push_back(insig);
  };
  long long int black = 0;
  long long int white = 0;
  for (long long int i = 0; i < N; i++) {
    if (i % 2 == 0) {
      black += ((a[i] + 1) / 2);
      white += (a[i] / 2);
    } else {
      black += ((a[i]) / 2);
      white += ((a[i] + 1) / 2);
    }
  }
  cout << min(black, white) << endl;
  return 0;
}

### 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;
long long int insig;
void Out(vector<long long int> x) {
  for (long long int i = 0; i < x.size(); i++) {
    cout << x[i] << " ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int N;
  cin >> N;
  vector<long long int> a;
  for (long long int IBRO = 0; IBRO < N; IBRO++) {
    cin >> insig;
    a.push_back(insig);
  };
  long long int black = 0;
  long long int white = 0;
  for (long long int i = 0; i < N; i++) {
    if (i % 2 == 0) {
      black += ((a[i] + 1) / 2);
      white += (a[i] / 2);
    } else {
      black += ((a[i]) / 2);
      white += ((a[i] + 1) / 2);
    }
  }
  cout << min(black, white) << endl;
  return 0;
}

```