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;
long long mod = 1000000007;
template <class C>
void mini(C& a, C b) {
  a = min(a, b);
}
template <class C>
void maxi(C& a, C b) {
  a = max(a, b);
}
void a() {
  int n, k;
  cin >> n >> k;
  string s;
  cin >> s;
  vector<int> in(n);
  for (int i = 0; i < n; i++) in[i] = (int)s[i] - (int)'0';
  cout << n << '\n';
  vector<int> out(n);
  int i = 0;
  while (i < k) {
    out[i] = in[i];
    i++;
  }
  while (i < n) {
    out[i] = out[i - k];
    i++;
  }
  if (out >= in) {
    for (int i = 0; i < n; i++) cout << out[i];
    cout << '\n';
    return;
  }
  i = k - 1;
  while (out[i] == 9) {
    out[i] = 0;
    i--;
  }
  out[i]++;
  i = k;
  while (i < n) {
    out[i] = out[i - k];
    i++;
  }
  for (int i = 0; i < n; i++) cout << out[i];
  cout << '\n';
}
void b() {
  int n;
  cin >> n;
  vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  long long blancas = 0;
  long long negras = 0;
  for (int i = 0; i < n; i++) {
    blancas += a[i] / 2;
    negras += a[i] / 2;
    if (a[i] % 2 == 1) {
      if (i % 2 == 0)
        blancas++;
      else
        negras++;
    }
  }
  cout << min(blancas, negras) << '\n';
}
int main() {
  b();
  return 0;
}
