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 long long inf = 2000000000000000000LL;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t, i, j, k, l, m, n, o, p, q, temp, ans = 0, flag = 0,
                                                mod = 1000000007;
  cin >> n;
  map<long long, long long> mp;
  for (i = 0; i < n; i++) {
    cin >> temp;
    mp[temp]++;
  }
  vector<long long> v;
  for (auto it = mp.begin(); it != mp.end(); it++) {
    ans += ((it->first) * ((it->second) / 2));
    v.push_back((it->first) * ((it->second) % 2));
  }
  sort(v.begin(), v.end(), greater<int>());
  long long black = 0, white = 0;
  for (i = 0; i < v.size(); i++) {
    if (i % 2) {
      black += ((1 + v[i]) / 2);
      white += (v[i] / 2);
    } else {
      black += (v[i] / 2);
      white += ((1 + v[i]) / 2);
    }
  }
  cout << ans + min(black, white);
}
