#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const double PI = acos(-1);
const long long MOD = 1000000007;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<long long> a(N);
  for (int i = 0; i < (int)(N); i++) cin >> a[i];
  long long p, q;
  p = q = 0;
  for (int i = 0; i < (int)(N); i++) {
    if (i % 2 == 0) {
      p += ((a[i] + 1LL) / 2LL);
      q += (a[i] / 2LL);
    } else {
      p += (a[i] / 2LL);
      q += ((a[i] + 1LL) / 2LL);
    }
  }
  cout << min(p, q) << endl;
}

### Prompt

Create a solution in cpp for 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>
#pragma GCC optimize("Ofast")
using namespace std;
const double PI = acos(-1);
const long long MOD = 1000000007;
template <class T>
inline bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  vector<long long> a(N);
  for (int i = 0; i < (int)(N); i++) cin >> a[i];
  long long p, q;
  p = q = 0;
  for (int i = 0; i < (int)(N); i++) {
    if (i % 2 == 0) {
      p += ((a[i] + 1LL) / 2LL);
      q += (a[i] / 2LL);
    } else {
      p += (a[i] / 2LL);
      q += ((a[i] + 1LL) / 2LL);
    }
  }
  cout << min(p, q) << endl;
}

```