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;
template <typename... T>
void read(T&... args) {
  ((cin >> args), ...);
}
long long int add(long long int a, long long int b) {
  a %= 998244353;
  b %= 998244353;
  return (a + b + 998244353) % 998244353;
}
long long int mul(long long int a, long long int b) {
  a %= 998244353;
  b %= 998244353;
  return (a * b + 998244353) % 998244353;
}
long long int power(long long int x, long long int n) {
  if (n == 0 || x == 1) return 1;
  long long int ans = power(x, n / 2) % 998244353;
  if (n & 1)
    return (((ans * ans) % 998244353) * (x % 998244353)) % 998244353;
  else
    return ((ans * ans) % 998244353);
}
void __solve() {
  long long int n;
  read(n);
  long long int a[n + 1];
  a[0] = 0;
  for (long long int i = 0; i < n; i++) read(a[i + 1]);
  long long int w = 0, b = 0;
  for (long long int i = 1; i <= n; i++) {
    if (i % 2) {
      w += a[i] / 2 + a[i] % 2;
      b += a[i] / 2;
    } else {
      b += a[i] / 2 + a[i] % 2;
      w += a[i] / 2;
    }
  }
  cout << min(w, b);
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int tc;
  tc = 1;
  while (tc--) {
    __solve();
  }
  return 0;
}
