#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const long long maxn = 3e5 + 10;
long long n;
long long a[maxn];
long long dp[maxn];
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  long long ans = 0;
  long long t1 = 0, t2 = 0;
  for (int i = 1; i <= n; ++i) {
    ans += a[i] / 2;
    if (a[i] & 1) {
      if (i % 2 == 0)
        t1++;
      else
        t2++;
    }
  }
  ans += min(t1, t2);
  cout << ans << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
inline long long read() {
  char ch = getchar();
  long long x = 0, f = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * f;
}
const long long maxn = 3e5 + 10;
long long n;
long long a[maxn];
long long dp[maxn];
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  n = read();
  for (int i = 1; i <= n; ++i) a[i] = read();
  long long ans = 0;
  long long t1 = 0, t2 = 0;
  for (int i = 1; i <= n; ++i) {
    ans += a[i] / 2;
    if (a[i] & 1) {
      if (i % 2 == 0)
        t1++;
      else
        t2++;
    }
  }
  ans += min(t1, t2);
  cout << ans << endl;
  return 0;
}

```