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 int pow(long long int x, long long int y, long long int m) {
  long long int res = 1;
  x = x % m;
  while (y > 0) {
    if (y & 1) res = ((res % m) * (x % m)) % m;
    y = y >> 1;
    x = ((x % m) * (x % m)) % m;
  }
  return res % m;
}
template <typename Arg1>
void ZZ(const char* name, Arg1&& arg1) {
  std::cerr << name << " = " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void ZZ(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " = " << arg1;
  ZZ(comma, args...);
}
const long long int INF = 0xFFFFFFFFFFFFFFFL;
clock_t time_p = clock();
void abhigyan10() {
  time_p = clock() - time_p;
  cerr << "Time Taken : " << (float)(time_p) / CLOCKS_PER_SEC << "\n";
}
long long int seed;
mt19937 rng(seed = chrono::steady_clock::now().time_since_epoch().count());
inline long long int rnd(long long int l = 0, long long int r = INF) {
  return uniform_int_distribution<long long int>(l, r)(rng);
}
const long long int mod = 1000000007;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, i, w = 0, b = 0;
  cin >> n;
  vector<int> a(n);
  for (i = 0; i < n; i++) {
    cin >> a[i];
    w += a[i] / 2 + (a[i] % 2) * ((i + 1) % 2);
    b += a[i] / 2 + (a[i] % 2) * (i % 2);
  }
  cout << min(w, b);
  return 0;
}
