#include <bits/stdc++.h>
using namespace std;
template <typename elemType>
inline void Read(elemType &T) {
  elemType X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  T = (w ? -X : X);
}
long long Data[300010];
int N;
int main() {
  Read(N);
  long long A = 0, B = 0;
  for (register int i = 1; i <= N; ++i) {
    Read(Data[i]);
    if (i & 1) {
      A += Data[i] >> 1;
      B += Data[i] - (Data[i] >> 1);
    } else {
      A += Data[i] - (Data[i] >> 1);
      B += Data[i] >> 1;
    }
  }
  printf("%I64d\n", min(A, B));
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
template <typename elemType>
inline void Read(elemType &T) {
  elemType X = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
  T = (w ? -X : X);
}
long long Data[300010];
int N;
int main() {
  Read(N);
  long long A = 0, B = 0;
  for (register int i = 1; i <= N; ++i) {
    Read(Data[i]);
    if (i & 1) {
      A += Data[i] >> 1;
      B += Data[i] - (Data[i] >> 1);
    } else {
      A += Data[i] - (Data[i] >> 1);
      B += Data[i] >> 1;
    }
  }
  printf("%I64d\n", min(A, B));
  return 0;
}

```