#include <bits/stdc++.h>
using namespace std;
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
  return s;
}
mt19937 rnd(time(NULL));
const int maxN = 3e5 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
          bs = 29;
long long arr[maxN], ans;
int n;
long long F(long long n) { return n * (n + 1) / 2; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int hei = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long prof = (arr[i] - hei) / 2;
    ans += prof, arr[i] -= prof * 2;
    if (hei && arr[i] == hei)
      ans += hei, hei--;
    else
      hei = arr[i];
  }
  long long tmp = F(hei / 2) * 2;
  ans += min(F(hei) - tmp, tmp);
  cout << ans << '\n';
}

### Prompt

Generate 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;
class DebugStream {
} LOG;
template <typename T>
DebugStream &operator<<(DebugStream &s, const T &) {
  return s;
}
mt19937 rnd(time(NULL));
const int maxN = 3e5 + 9, maxV = 1e6 + 9, MOD = 1e9 + 7, SQ = 335, lg = 20,
          bs = 29;
long long arr[maxN], ans;
int n;
long long F(long long n) { return n * (n + 1) / 2; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  int hei = 0;
  for (int i = n - 1; i >= 0; i--) {
    long long prof = (arr[i] - hei) / 2;
    ans += prof, arr[i] -= prof * 2;
    if (hei && arr[i] == hei)
      ans += hei, hei--;
    else
      hei = arr[i];
  }
  long long tmp = F(hei / 2) * 2;
  ans += min(F(hei) - tmp, tmp);
  cout << ans << '\n';
}

```