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 int N = (int)3e5;
long long int mpow(long long int a, long long int b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    long long int memo = mpow(a, b / 2);
    return memo * memo % (long long int)(1e9 + 7);
  } else
    return mpow(a, b - 1) * a % (long long int)(1e9 + 7);
}
long long int lpow(long long int a, long long int b) {
  if (b == 0)
    return 1;
  else if (b % 2 == 0) {
    long long int memo = lpow(a, b / 2);
    return memo * memo;
  } else
    return lpow(a, b - 1) * a;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
vector<long long int> kaijo_memo;
long long int kaijo(long long int n) {
  if (kaijo_memo.size() > n) return kaijo_memo[n];
  if (kaijo_memo.size() == 0) kaijo_memo.push_back(1);
  while (kaijo_memo.size() <= n)
    kaijo_memo.push_back(kaijo_memo[kaijo_memo.size() - 1] * kaijo_memo.size() %
                         (long long int)(1e9 + 7));
  return kaijo_memo[n];
}
vector<long long int> gyaku_kaijo_memo;
long long int gyaku_kaijo(long long int n) {
  if (gyaku_kaijo_memo.size() > n) return gyaku_kaijo_memo[n];
  if (gyaku_kaijo_memo.size() == 0) gyaku_kaijo_memo.push_back(1);
  while (gyaku_kaijo_memo.size() <= n)
    gyaku_kaijo_memo.push_back(
        gyaku_kaijo_memo[gyaku_kaijo_memo.size() - 1] *
        mpow(gyaku_kaijo_memo.size(), (long long int)(1e9 + 7) - 2) %
        (long long int)(1e9 + 7));
  return gyaku_kaijo_memo[n];
}
long long int nCr(long long int n, long long int r) {
  if (n == r) return 1;
  if (n < r || r < 0) return 0;
  long long int ret = 1;
  ret *= kaijo(n);
  ret %= (long long int)(1e9 + 7);
  ret *= gyaku_kaijo(r);
  ret %= (long long int)(1e9 + 7);
  ret *= gyaku_kaijo(n - r);
  ret %= (long long int)(1e9 + 7);
  return ret;
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<long long int> A;
  for (int i = 0; i < (int)(n); i++) {
    long long int a;
    cin >> a;
    A.push_back(a);
  }
  reverse(A.begin(), A.end());
  int now = 0;
  long long int ans = 0;
  long long int x = 0, y = 0;
  for (int i = 0; i < (int)(n); i++) {
    long long int a = A[i];
    x += a / 2 + (i % 2 == 0 ? 0 : a % 2);
    y += a / 2 + (i % 2 == 1 ? 0 : a % 2);
  }
  cout << min(x, y) << endl;
  return 0;
}
