#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T>
void prContain(const T& x) {
  pr("{");
  bool fst = 1;
  for (auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
template <typename T>
int remin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
int remax(T& a, const T& b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
int n;
int main(int argc, const char** argv) {
  scanf("%d", &n);
  vector<int> c(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
  }
  long long w = 0, b = 0;
  for (int i = 0; i < n; ++i) {
    if (i & 1) {
      w += c[i] / 2;
      b += (c[i] + 1) / 2;
    } else {
      w += (c[i] + 1) / 2;
      b += c[i] / 2;
    }
  }
  printf("%lld\n", min(w, b));
  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;
template <class T1, class T2>
void pr(const pair<T1, T2>& x);
template <class T, size_t SZ>
void pr(const array<T, SZ>& x);
template <class T>
void pr(const vector<T>& x);
template <class T>
void pr(const set<T>& x);
template <class T1, class T2>
void pr(const map<T1, T2>& x);
template <class T>
void pr(const T& x) {
  cout << x;
}
template <class Arg, class... Args>
void pr(const Arg& first, const Args&... rest) {
  pr(first);
  pr(rest...);
}
template <class T1, class T2>
void pr(const pair<T1, T2>& x) {
  pr("{", x.first, ", ", x.second, "}");
}
template <class T>
void prContain(const T& x) {
  pr("{");
  bool fst = 1;
  for (auto& a : x) pr(!fst ? ", " : "", a), fst = 0;
  pr("}");
}
template <class T, size_t SZ>
void pr(const array<T, SZ>& x) {
  prContain(x);
}
template <class T>
void pr(const vector<T>& x) {
  prContain(x);
}
template <class T>
void pr(const set<T>& x) {
  prContain(x);
}
template <class T1, class T2>
void pr(const map<T1, T2>& x) {
  prContain(x);
}
void ps() { pr("\n"); }
template <class Arg, class... Args>
void ps(const Arg& first, const Args&... rest) {
  pr(first, " ");
  ps(rest...);
}
template <typename T>
int remin(T& a, const T& b) {
  if (b < a) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
int remax(T& a, const T& b) {
  if (b > a) {
    a = b;
    return true;
  }
  return false;
}
int n;
int main(int argc, const char** argv) {
  scanf("%d", &n);
  vector<int> c(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
  }
  long long w = 0, b = 0;
  for (int i = 0; i < n; ++i) {
    if (i & 1) {
      w += c[i] / 2;
      b += (c[i] + 1) / 2;
    } else {
      w += (c[i] + 1) / 2;
      b += c[i] / 2;
    }
  }
  printf("%lld\n", min(w, b));
  return 0;
}

```