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;
int mod = (ll)1e9 + 7;
const int INF = 1e9 + 1;
const int N = 2e5 + 100;
const double eps = 1e-7;
template <class T>
using V = vector<T>;
template <class T>
using VV = V<V<T>>;
template <typename XPAX>
void ckma(XPAX &x, XPAX y) {
  x = (x < y ? y : x);
}
template <typename XPAX>
void ckmi(XPAX &x, XPAX y) {
  x = (x > y ? y : x);
}
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void solve() {
  int n;
  cin >> n;
  V<int> a(n);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  V<ll> cnt(2, 0);
  for (int i = 0; i < (n); ++i) {
    cnt[i % 2] += a[i] / 2;
    cnt[(i + 1) % 2] += (a[i] + 1) / 2;
  }
  cerr << "["
       << "cnt[0], cnt[1]"
       << "] = [";
  _print(cnt[0], cnt[1]);
  cout << min(cnt[0], cnt[1]) << '\n';
}
void test_case() {
  int t;
  cin >> t;
  for (int p = 0; p < (t); ++p) {
    solve();
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  solve();
}
