#include <bits/stdc++.h>
using namespace std;
const int maxn = 3e5 + 5;
int temp[maxn];
int Id_temp;
long long sum;
long long sumB, sumW;
inline void dec2() {
  sum += temp[Id_temp];
  Id_temp--;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    if (Id_temp && t == temp[Id_temp])
      dec2();
    else
      temp[++Id_temp] = t;
  }
  for (int i = 1, ii = 1; i <= Id_temp; i++, ii ^= 1) {
    if (ii)
      sumB += temp[i] / 2, sumW += temp[i] - (temp[i] / 2);
    else
      sumW += temp[i] / 2, sumB += temp[i] - (temp[i] / 2);
  }
  sum += min(sumB, sumW);
  printf("%lld\n", sum);
}

### Prompt

Please formulate 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 int maxn = 3e5 + 5;
int temp[maxn];
int Id_temp;
long long sum;
long long sumB, sumW;
inline void dec2() {
  sum += temp[Id_temp];
  Id_temp--;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int t;
    scanf("%d", &t);
    if (Id_temp && t == temp[Id_temp])
      dec2();
    else
      temp[++Id_temp] = t;
  }
  for (int i = 1, ii = 1; i <= Id_temp; i++, ii ^= 1) {
    if (ii)
      sumB += temp[i] / 2, sumW += temp[i] - (temp[i] / 2);
    else
      sumW += temp[i] / 2, sumB += temp[i] - (temp[i] / 2);
  }
  sum += min(sumB, sumW);
  printf("%lld\n", sum);
}

```