#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
void amax(long long& a, long long b) { a = max(a, b) % 1000000007; }
void amin(long long& a, long long b) { a = min(a, b) % 1000000007; }
void inc(long long& a, long long b) { a = (a + b) % 1000000007; }
void dec(long long& a, long long b) { a = (a - b + 1000000007) % 1000000007; }
long long solve() {
  long long black = 0, white = 0;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n; i++) {
    if (i % 2) {
      white += (a[i] + 1) / 2;
      black += a[i] / 2;
    } else {
      white += a[i] / 2;
      black += (a[i] + 1) / 2;
    }
  }
  cout << min(white, black);
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a CPP coded solution 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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse2")
using namespace std;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long max(long long a, long long b) { return (a > b) ? a : b; }
long long min(long long a, long long b) { return (a < b) ? a : b; }
void amax(long long& a, long long b) { a = max(a, b) % 1000000007; }
void amin(long long& a, long long b) { a = min(a, b) % 1000000007; }
void inc(long long& a, long long b) { a = (a + b) % 1000000007; }
void dec(long long& a, long long b) { a = (a - b + 1000000007) % 1000000007; }
long long solve() {
  long long black = 0, white = 0;
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  for (long long i = 0; i < n; i++) {
    if (i % 2) {
      white += (a[i] + 1) / 2;
      black += a[i] / 2;
    } else {
      white += a[i] / 2;
      black += (a[i] + 1) / 2;
    }
  }
  cout << min(white, black);
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
}

```