#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int n, a[300010];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long long ans = 0;
  int pre = 0, odd = 0;
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j + 1 <= n && a[j + 1] == a[i]) j++;
    int L = j - i + 1;
    if (L & 1) {
      if (pre == 0) {
        ans += 1LL * a[i] * L / 2;
        if (a[i] & 1) {
          pre = 1;
          odd = 1;
        }
      } else {
        int top = 0;
        if (odd)
          top = 2 * pre - 1;
        else
          top = 2 * pre;
        while (top > a[i]) {
          top -= 2;
          pre--;
        }
        while (a[i] - 2 >= top) {
          ans += 1LL * L;
          a[i] -= 2;
        }
        ans += 1LL * a[i] * (L / 2) + pre;
        if (odd) {
          if (top == a[i]) pre--;
        } else {
          if (top < a[i]) pre++;
        }
        odd ^= 1;
      }
    } else {
      ans += 1LL * a[i] * L / 2;
    }
    i = j;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int n, a[300010];
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  long long ans = 0;
  int pre = 0, odd = 0;
  for (int i = 1; i <= n; ++i) {
    int j = i;
    while (j + 1 <= n && a[j + 1] == a[i]) j++;
    int L = j - i + 1;
    if (L & 1) {
      if (pre == 0) {
        ans += 1LL * a[i] * L / 2;
        if (a[i] & 1) {
          pre = 1;
          odd = 1;
        }
      } else {
        int top = 0;
        if (odd)
          top = 2 * pre - 1;
        else
          top = 2 * pre;
        while (top > a[i]) {
          top -= 2;
          pre--;
        }
        while (a[i] - 2 >= top) {
          ans += 1LL * L;
          a[i] -= 2;
        }
        ans += 1LL * a[i] * (L / 2) + pre;
        if (odd) {
          if (top == a[i]) pre--;
        } else {
          if (top < a[i]) pre++;
        }
        odd ^= 1;
      }
    } else {
      ans += 1LL * a[i] * L / 2;
    }
    i = j;
  }
  cout << ans << endl;
  return 0;
}

```