#include <bits/stdc++.h>
using namespace std;
const long long INF = 0x3f3f3f3f;
const long long inf = pow(10, 18);
long long modulo = pow(10, 9) + 7;
ifstream in;
ofstream out;
void deal() {
  long long n;
  cin >> n;
  vector<long long> all(n);
  long long last = 0;
  long long s = 0;
  for (long long i = 0; i < n; i++) {
    cin >> all[i];
    long long last_oddity = (last % 2);
    long long cur_oddity = (all[i] % 2);
    if (last > all[i]) {
      if (last_oddity == cur_oddity) {
        last = all[i];
      } else {
        last = all[i] - 1;
      }
    }
    s += (last + 1) / 2;
    s += (all[i] - last) / 2;
    if (last_oddity == cur_oddity) {
      if (last != 0) {
        --last;
      }
    } else {
      ++last;
    }
  }
  cout << s;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  deal();
}

### 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 INF = 0x3f3f3f3f;
const long long inf = pow(10, 18);
long long modulo = pow(10, 9) + 7;
ifstream in;
ofstream out;
void deal() {
  long long n;
  cin >> n;
  vector<long long> all(n);
  long long last = 0;
  long long s = 0;
  for (long long i = 0; i < n; i++) {
    cin >> all[i];
    long long last_oddity = (last % 2);
    long long cur_oddity = (all[i] % 2);
    if (last > all[i]) {
      if (last_oddity == cur_oddity) {
        last = all[i];
      } else {
        last = all[i] - 1;
      }
    }
    s += (last + 1) / 2;
    s += (all[i] - last) / 2;
    if (last_oddity == cur_oddity) {
      if (last != 0) {
        --last;
      }
    } else {
      ++last;
    }
  }
  cout << s;
}
int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(0);
  deal();
}

```