#include <bits/stdc++.h>
using namespace std;
inline long long bip(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * x) % mod;
    }
    x = (x * x) % mod;
    n >>= 1;
  }
  return res;
}
const int maxn = 1e9;
const long long inf = 1e18;
const int mod = 1e9 + 7;
const int N = 3e5 + 10;
int n, x;
void solve() {
  cin >> n;
  long long w = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (i & 1) {
      w += (x + 1) / 2;
      b += x / 2;
    } else {
      w += x / 2;
      b += (x + 1) / 2;
    }
  }
  cout << min(w, b);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
    cout << '\n';
  }
}

### 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;
inline long long bip(long long x, long long n, long long mod) {
  long long res = 1;
  while (n) {
    if (n & 1) {
      res = (res * x) % mod;
    }
    x = (x * x) % mod;
    n >>= 1;
  }
  return res;
}
const int maxn = 1e9;
const long long inf = 1e18;
const int mod = 1e9 + 7;
const int N = 3e5 + 10;
int n, x;
void solve() {
  cin >> n;
  long long w = 0, b = 0;
  for (int i = 1; i <= n; i++) {
    cin >> x;
    if (i & 1) {
      w += (x + 1) / 2;
      b += x / 2;
    } else {
      w += x / 2;
      b += (x + 1) / 2;
    }
  }
  cout << min(w, b);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
    cout << '\n';
  }
}

```