#include <bits/stdc++.h>
using namespace std;
const int mx2 = 505;
const int mx3 = 1005;
const int mx5 = 300005;
const int mx6 = 800005;
int _toggle(int N, int pos) { return N = N ^ (1 << pos); }
int _set(int N, int pos) { return N = N | (1 << pos); }
int _reset(int N, int pos) { return N = N & ~(1 << pos); }
bool _check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool _upper(char a) { return a >= 'A' && a <= 'Z'; }
bool _lower(char a) { return a >= 'a' && a <= 'z'; }
bool _digit(char a) { return a >= '0' && a <= '9'; }
int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[] = {0, 0, 1, -1, -1, 1, -1, 1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a;
  long long w = 0, b = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    int x = a / 2, y = (a + 1) / 2;
    if (i & 1) {
      w += y, b += x;
    } else {
      w += x, b += y;
    }
  }
  cout << min(w, b) << '\n';
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 mx2 = 505;
const int mx3 = 1005;
const int mx5 = 300005;
const int mx6 = 800005;
int _toggle(int N, int pos) { return N = N ^ (1 << pos); }
int _set(int N, int pos) { return N = N | (1 << pos); }
int _reset(int N, int pos) { return N = N & ~(1 << pos); }
bool _check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool _upper(char a) { return a >= 'A' && a <= 'Z'; }
bool _lower(char a) { return a >= 'a' && a <= 'z'; }
bool _digit(char a) { return a >= '0' && a <= '9'; }
int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[] = {0, 0, 1, -1, -1, 1, -1, 1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, a;
  long long w = 0, b = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a;
    int x = a / 2, y = (a + 1) / 2;
    if (i & 1) {
      w += y, b += x;
    } else {
      w += x, b += y;
    }
  }
  cout << min(w, b) << '\n';
}

```