#include <bits/stdc++.h>
using namespace std;
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 3e5 + 15;
long long dp[N][2];
int main() {
  int n;
  cin >> n;
  int a[n + 2];
  long long b = 0, w = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i & 1) {
      b += (a[i] + 1) / 2;
      w += a[i] / 2;
    } else {
      w += (a[i] + 1) / 2;
      b += a[i] / 2;
    }
  }
  cout << min(b, w) << "\n";
  return 0;
}

### Prompt

Please create a solution in CPP 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;
template <typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
  cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char *names, Arg1 &&arg1, Args &&...args) {
  const char *comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const int N = 3e5 + 15;
long long dp[N][2];
int main() {
  int n;
  cin >> n;
  int a[n + 2];
  long long b = 0, w = 0;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i & 1) {
      b += (a[i] + 1) / 2;
      w += a[i] / 2;
    } else {
      w += (a[i] + 1) / 2;
      b += a[i] / 2;
    }
  }
  cout << min(b, w) << "\n";
  return 0;
}

```