#include <bits/stdc++.h>
const long long maxn = 3e6 + 9;
long long Read() {
  long long x(0), f(1);
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3ll) + (x << 1ll) + c - '0';
    c = getchar();
  }
  return x * f;
}
long long T, n;
long long a[maxn], sum[2];
int main() {
  n = Read();
  for (long long i = 1; i <= n; ++i) {
    long long x(Read());
    if ((x + i) & 1) {
      sum[1] += (x + 1) / 2;
      sum[0] += (x - (x + 1) / 2);
    } else {
      sum[0] += (x + 1) / 2;
      sum[1] += (x - (x + 1) / 2);
    }
  }
  std::cout << std::min(sum[0], sum[1]);
}

### Prompt

Please create a solution in cpp 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>
const long long maxn = 3e6 + 9;
long long Read() {
  long long x(0), f(1);
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3ll) + (x << 1ll) + c - '0';
    c = getchar();
  }
  return x * f;
}
long long T, n;
long long a[maxn], sum[2];
int main() {
  n = Read();
  for (long long i = 1; i <= n; ++i) {
    long long x(Read());
    if ((x + i) & 1) {
      sum[1] += (x + 1) / 2;
      sum[0] += (x - (x + 1) / 2);
    } else {
      sum[0] += (x + 1) / 2;
      sum[1] += (x - (x + 1) / 2);
    }
  }
  std::cout << std::min(sum[0], sum[1]);
}

```