#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
long long n;
long long a[maxn];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) cin >> a[i], sum += a[i];
  if (n & 1) n++;
  vector<long long> v;
  for (long long i = 1; i <= n; i += 2) {
    long long mask = ((a[i] & 1) << 1) + (a[i + 1] & 1);
    if (mask == 0 || mask == 3)
      continue;
    else if (((long long)(v).size()) > 0) {
      if (v.back() != mask)
        v.pop_back();
      else
        v.push_back(mask);
    } else
      v.push_back(mask);
  }
  cout << sum / 2 - ((long long)(v).size()) / 2 << '\n';
}

### Prompt

In cpp, your task is to solve 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;
const int maxn = 1e6 + 7;
const int inf = INT_MAX;
const long long inff = 1e18;
const long long mod = 1e9 + 7;
long long n;
long long a[maxn];
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  long long sum = 0;
  for (long long i = 1; i <= n; i++) cin >> a[i], sum += a[i];
  if (n & 1) n++;
  vector<long long> v;
  for (long long i = 1; i <= n; i += 2) {
    long long mask = ((a[i] & 1) << 1) + (a[i + 1] & 1);
    if (mask == 0 || mask == 3)
      continue;
    else if (((long long)(v).size()) > 0) {
      if (v.back() != mask)
        v.pop_back();
      else
        v.push_back(mask);
    } else
      v.push_back(mask);
  }
  cout << sum / 2 - ((long long)(v).size()) / 2 << '\n';
}

```