#include <bits/stdc++.h>
using namespace std;
long long heights[300001], check[300001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int columns;
  cin >> columns;
  for (int i{1}; i <= columns; ++i) cin >> heights[i];
  long long output = 0;
  vector<long long> basic;
  memset(check, 0, sizeof(check));
  for (int i{1}; i < columns; ++i) {
    if (heights[i] == heights[i + 1]) {
      output += heights[i];
      check[i] = check[i + 1] = 1;
      ++i;
    } else
      basic.push_back(heights[i]);
  }
  if (!check[columns]) basic.push_back(heights[columns]);
  long long white = 0, black = 0;
  int p = 0;
  for (int i{0}; i < basic.size(); ++i) {
    if (p) {
      black += basic[i] / 2;
      white += basic[i] - basic[i] / 2;
    } else {
      white += basic[i] / 2;
      black += basic[i] - basic[i] / 2;
    }
    p = !p;
  }
  cout << output + min(white, black);
  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;
long long heights[300001], check[300001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int columns;
  cin >> columns;
  for (int i{1}; i <= columns; ++i) cin >> heights[i];
  long long output = 0;
  vector<long long> basic;
  memset(check, 0, sizeof(check));
  for (int i{1}; i < columns; ++i) {
    if (heights[i] == heights[i + 1]) {
      output += heights[i];
      check[i] = check[i + 1] = 1;
      ++i;
    } else
      basic.push_back(heights[i]);
  }
  if (!check[columns]) basic.push_back(heights[columns]);
  long long white = 0, black = 0;
  int p = 0;
  for (int i{0}; i < basic.size(); ++i) {
    if (p) {
      black += basic[i] / 2;
      white += basic[i] - basic[i] / 2;
    } else {
      white += basic[i] / 2;
      black += basic[i] - basic[i] / 2;
    }
    p = !p;
  }
  cout << output + min(white, black);
  return 0;
}

```