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 maxn = 1e6 + 9;
const int N = 1 << 20;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
const int temp = 233;
const double eps = 0.0000001;
const double PI = acos(-1);
const int dx[] = {0, 0, 1, 1, 1, -1, -1, -1};
const int dy[] = {-1, 1, 1, -1, 0, 1, -1, 0};
inline int read() {
  int num = 0, w = 0;
  char ch = 0;
  while (!isdigit(ch)) {
    w |= ch == '-';
    ch = getchar();
  }
  while (isdigit(ch)) {
    num = (num << 3) + (num << 1) + (ch ^ 48);
    ch = getchar();
  }
  return w ? -num : num;
}
long long ans = 0;
int n;
long long a[maxn];
int main() {
  scanf("%d", &n);
  int t = 0;
  long long x = 0, y = 0;
  ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    x += a[i] / 2;
    y += a[i] / 2;
    if (a[i] % 2 == 1) {
      if (i % 2 == 1)
        x++;
      else
        y++;
    }
    t = 1 - t;
  }
  ans = min(x, y);
  printf("%lld\n", ans);
  return 0;
}
