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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
inline void INP();
const long long N = 3e6 + 7;
const long long INF = LLONG_MAX;
const long long inf = INT_MAX;
long long n, Q;
long long arr[N];
void solve() {
  cin >> n;
  long long ans = 0;
  for (long long i = 0; i < (long long)n; i++) {
    cin >> arr[i];
  }
  vector<long long> vec;
  for (long long i = 0; i < (long long)n; i++) {
    if (i < n and arr[i] == arr[i + 1])
      ans += arr[i], i++;
    else
      vec.push_back(arr[i]);
  }
  long long b = 0, w = 0, f = 1;
  for (long long i = 0; i < (long long)vec.size(); i++) {
    w += vec[i] / 2;
    b += vec[i] / 2;
    if (vec[i] & 1) {
      if (f)
        w++;
      else
        b++;
    }
    f = !f;
  }
  cout << ans + min(w, b) << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  INP();
  long long t = 1;
  while (t--) solve();
}
inline void INP() {}
