#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  long long ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
long long add(long long x, long long y) {
  return x + y > 998244353 ? x + y - 998244353 : x + y;
}
long long sub(long long x, long long y) {
  return x - y < 0 ? x - y + 998244353 : x - y;
}
long long a[300005];
signed main() {
  long long n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  long long r = 0, b = 0;
  for (long long i = 1; i <= n; i++) {
    if (i & 1)
      r += a[i] / 2, b += (a[i] + 1) / 2;
    else
      b += a[i] / 2, r += (a[i] + 1) / 2;
  }
  cout << min(r, b) << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
inline long long read() {
  char ch = getchar();
  long long nega = 1;
  while (!isdigit(ch)) {
    if (ch == '-') nega = -1;
    ch = getchar();
  }
  long long ans = 0;
  while (isdigit(ch)) {
    ans = ans * 10 + ch - 48;
    ch = getchar();
  }
  if (nega == -1) return -ans;
  return ans;
}
long long add(long long x, long long y) {
  return x + y > 998244353 ? x + y - 998244353 : x + y;
}
long long sub(long long x, long long y) {
  return x - y < 0 ? x - y + 998244353 : x - y;
}
long long a[300005];
signed main() {
  long long n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  long long r = 0, b = 0;
  for (long long i = 1; i <= n; i++) {
    if (i & 1)
      r += a[i] / 2, b += (a[i] + 1) / 2;
    else
      b += a[i] / 2, r += (a[i] + 1) / 2;
  }
  cout << min(r, b) << endl;
  return 0;
}

```