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 N = 3e5 + 10;
int n, a[N], hole[N], type[N];
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long long ans = 0;
  hole[0] = type[0] = 0;
  for (int i = 1; i <= n; ++i)
    if (hole[i - 1] == 0) {
      if (a[i] % 2 == 0) {
        ans += a[i] / 2;
        hole[i] = 0;
      } else {
        ans += a[i] / 2;
        hole[i] = 1;
        type[i] = 1;
      }
    } else if (type[i - 1] == 1) {
      type[i] = 0;
      ans += (a[i] + 1) / 2;
      if ((a[i] + 1) / 2 <= hole[i - 1])
        hole[i] = (a[i] % 2 == 1 ? ((a[i] + 1) / 2 - 1) : (a[i] + 1) / 2);
      else
        hole[i] = (a[i] % 2 == 1 ? hole[i - 1] - 1 : hole[i - 1]);
    } else {
      type[i] = 1;
      ans += a[i] / 2;
      if (a[i] / 2 <= hole[i - 1])
        hole[i] = (a[i] % 2 == 1 ? a[i] / 2 + 1 : a[i] / 2);
      else
        hole[i] = (a[i] % 2 == 1 ? hole[i - 1] + 1 : hole[i - 1]);
    }
  cout << ans << endl;
}
