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 a[300005], b[300005];
int main() {
  int n;
  cin >> n;
  int i, j;
  for (i = 0; i < 3e5 + 5; i++) b[i] = 0;
  long long ans = 0;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    b[1] += 1;
    b[1 + a[i]] -= 1;
  }
  long long s = 0;
  bool e = 0;
  for (i = 1; i <= n; i++) {
    if (s == 0) {
      ans += a[i] / 2;
      if (a[i] % 2) s = 1, e = 1;
    } else {
      ans += a[i] / 2;
      if (e) {
        if (a[i] % 2) {
          ans++;
          e = 0;
          s = min(s - 1, a[i] / 2);
        } else {
          e = 0;
          s = min(s, a[i] / 2);
        }
      } else {
        if (a[i] % 2) {
          e = 1;
          s = min(s, a[i] / 2) + 1;
        } else {
          e = 1;
          s = min(s, a[i] / 2);
        }
      }
    }
  }
  long long ans1 = ans;
  ans = 0;
  s = 0;
  for (i = 1; i <= a[1]; i++) b[i] += b[i - 1];
  s = 0;
  e = 0;
  for (i = 1; i <= a[i]; i++) {
    if (s == 0) {
      ans += b[i] / 2;
      if (b[i] % 2) s = 1, e = 1;
    } else {
      ans += b[i] / 2;
      if (e) {
        if (b[i] % 2) {
          ans++;
          e = 0;
          s = min(s - 1, b[i] / 2);
        } else {
          e = 0;
          s = min(s, b[i] / 2);
        }
      } else {
        if (a[i] % 2) {
          e = 1;
          s = min(s, b[i] / 2) + 1;
        } else {
          e = 1;
          s = min(s, b[i] / 2);
        }
      }
    }
  }
  cout << max(ans, ans1);
  return 0;
}
