#include <bits/stdc++.h>
using namespace std;
template <class t>
t mag(t a) {
  return a > 0 ? a : -a;
}
template <class T, class U>
T cdiv(T num, U den) {
  return (num / den) + ((num % den) > 0);
}
std::ifstream term("/dev/tty");
std::string tempstring;
using ll = int_fast64_t;
using ii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
const int sij = 400010;
const ll mod = 1000000007ll;
int n;
int esc(int tcase) {
  cin >> n;
  ll tot[2] = {0, 0};
  int cr = 0;
  for (auto i = 0; i != n; i++) {
    int tp;
    cin >> tp;
    tot[cr] += (tp / 2);
    tot[1 - cr] += tp - (tp / 2);
    cr = 1 - cr;
  }
  cout << min(tot[0], tot[1]);
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  for (int i = 1; i <= tc; i++) esc(i);
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
template <class t>
t mag(t a) {
  return a > 0 ? a : -a;
}
template <class T, class U>
T cdiv(T num, U den) {
  return (num / den) + ((num % den) > 0);
}
std::ifstream term("/dev/tty");
std::string tempstring;
using ll = int_fast64_t;
using ii = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vii = vector<ii>;
const int sij = 400010;
const ll mod = 1000000007ll;
int n;
int esc(int tcase) {
  cin >> n;
  ll tot[2] = {0, 0};
  int cr = 0;
  for (auto i = 0; i != n; i++) {
    int tp;
    cin >> tp;
    tot[cr] += (tp / 2);
    tot[1 - cr] += tp - (tp / 2);
    cr = 1 - cr;
  }
  cout << min(tot[0], tot[1]);
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tc = 1;
  for (int i = 1; i <= tc; i++) esc(i);
}

```