#include <bits/stdc++.h>
using namespace std;
int N;
int A[303030];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  signed long long C[2] = {};
  for (i = 0; i < (N); i++) {
    cin >> A[i];
    if (i % 2 == 0) {
      C[0] += A[i] / 2;
      C[1] += A[i] / 2 + A[i] % 2;
    } else {
      C[1] += A[i] / 2;
      C[0] += A[i] / 2 + A[i] % 2;
    }
  }
  cout << min(C[0], C[1]) << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int N;
int A[303030];
void solve() {
  int i, j, k, l, r, x, y;
  string s;
  cin >> N;
  signed long long C[2] = {};
  for (i = 0; i < (N); i++) {
    cin >> A[i];
    if (i % 2 == 0) {
      C[0] += A[i] / 2;
      C[1] += A[i] / 2 + A[i] % 2;
    } else {
      C[1] += A[i] / 2;
      C[0] += A[i] / 2 + A[i] % 2;
    }
  }
  cout << min(C[0], C[1]) << endl;
}
int main(int argc, char** argv) {
  string s;
  int i;
  if (argc == 1) ios::sync_with_stdio(false), cin.tie(0);
  for (i = 0; i < (argc - 1); i++) s += argv[i + 1], s += '\n';
  for (i = 0; i < (s.size()); i++) ungetc(s[s.size() - 1 - i], stdin);
  cout.tie(0);
  solve();
  return 0;
}

```