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;
using ll = long long;
using ld = long double;
const int mod = 1e9 + 7;
const int inf = 2e9 + 5;
const ll linf = 9e18 + 5;
int n;
const int N = 3e5 + 5;
ll arr[N];
ll ans = 0;
void init() {}
void input() {}
void solve() {
  cin >> n;
  for (int i = n - 1; i > -1; i--) {
    cin >> arr[i];
    ans += arr[i] / 2;
    arr[i] %= 2;
  }
  stack<int> nums;
  for (int i = 0; i < n; i++) {
    if (nums.empty()) {
      nums.push(arr[i]);
      continue;
    }
    if (arr[i] == nums.top()) {
      nums.pop();
      ans += arr[i];
    } else {
      nums.push(arr[i]);
    }
  }
  n = int(nums.size());
  for (int i = n - 1; i > -1; i--) {
    arr[i] = nums.top();
    nums.pop();
  }
  for (int i = 1; i < n; i++) {
    if (arr[i] == arr[i - 1] && arr[i] == 1) {
      arr[i] = 0;
      arr[i - 1] = 0;
      ans++;
    }
  }
  cout << ans;
}
void output() {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int number_of_testcases = 1;
  while (number_of_testcases--) {
    init();
    input();
    solve();
    output();
  }
}
