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>
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e6;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
bool compare(const pair<int, int>& i, const pair<int, int>& j) {
  if (i.first != j.first)
    return i.first < j.first;
  else
    return i.second < j.second;
}
int binarySearch(int arr[], int l, int r, int x) {
  if (r >= l) {
    int mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
    return binarySearch(arr, mid + 1, r, x);
  }
  return -1;
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  bool black = true;
  long long b = 0, w = 0;
  for (long long i = 0; i < n; i++) {
    if (black) {
      b += (a[i] / 2) + a[i] % 2;
      w += a[i] / 2;
    } else {
      b += a[i] / 2;
      w += (a[i] / 2) + a[i] % 2;
    }
    black = !black;
  }
  cout << min(b, w) << "\n";
}
