#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1e6 + 100;
const long long maxx = 1e13;
const long long mod1 = 1000000007;
const long long mod = 998244353;
const long long base = 31;
priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
               greater<pair<long long, long long>>>
    q;
vector<long long> adj[maxn];
vector<long long> adj1[maxn];
long long n, k, b;
long long a[12];
bool chk[maxn];
string ins(string s, long long a) {
  string b = s;
  b.push_back(char(a + '0'));
  return b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long sum = 0;
  bool y = false;
  long long wsum = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    wsum = wsum + (x + y) / 2;
    sum += x;
    y ^= 1;
  }
  cout << min(sum - wsum, wsum);
}

### 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;
const long long maxn = 1e6 + 100;
const long long maxx = 1e13;
const long long mod1 = 1000000007;
const long long mod = 998244353;
const long long base = 31;
priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
               greater<pair<long long, long long>>>
    q;
vector<long long> adj[maxn];
vector<long long> adj1[maxn];
long long n, k, b;
long long a[12];
bool chk[maxn];
string ins(string s, long long a) {
  string b = s;
  b.push_back(char(a + '0'));
  return b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long sum = 0;
  bool y = false;
  long long wsum = 0;
  for (int i = 1; i <= n; i++) {
    long long x;
    cin >> x;
    wsum = wsum + (x + y) / 2;
    sum += x;
    y ^= 1;
  }
  cout << min(sum - wsum, wsum);
}

```