#include <bits/stdc++.h>
using namespace std;
const long long sz = 200005;
long long mod = 1e9 + 7;
long long n, t, m;
vector<long long> BIT(sz, 0);
void Update(long long in, long long x) {
  for (long long i = in; i < sz; i += i & (-i)) BIT[i] += 1;
}
long long Query(long long in) {
  long long ans = 0;
  for (long long i = in; i >= 1; i -= i & (-i)) ans += BIT[i];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long> v(n);
  long long ans = 0;
  for (auto &i : v) cin >> i;
  long long i = 0;
  long long p = 0, q = 0;
  long long a1 = 0, a2 = 0;
  while (i < n) {
    p = v[i] / 2;
    q = v[i] - p;
    a1 += p;
    a2 += q;
    swap(a1, a2);
    i++;
  }
  cout << min(a1, a2) << endl;
  return 0;
}

### Prompt

Your challenge is to write a CPP solution 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;
const long long sz = 200005;
long long mod = 1e9 + 7;
long long n, t, m;
vector<long long> BIT(sz, 0);
void Update(long long in, long long x) {
  for (long long i = in; i < sz; i += i & (-i)) BIT[i] += 1;
}
long long Query(long long in) {
  long long ans = 0;
  for (long long i = in; i >= 1; i -= i & (-i)) ans += BIT[i];
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long> v(n);
  long long ans = 0;
  for (auto &i : v) cin >> i;
  long long i = 0;
  long long p = 0, q = 0;
  long long a1 = 0, a2 = 0;
  while (i < n) {
    p = v[i] / 2;
    q = v[i] - p;
    a1 += p;
    a2 += q;
    swap(a1, a2);
    i++;
  }
  cout << min(a1, a2) << endl;
  return 0;
}

```