#include <bits/stdc++.h>
using namespace std;
long long n, c1, c2, s[300300];
inline long long read() {
  long long w = 1, q = 0;
  char ch = ' ';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') q = q * 10 + ch - '0', ch = getchar();
  return w * q;
}
inline void write(long long x) {
  if (x < 0) {
    x = ~(x - 1);
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
signed main() {
  n = read();
  for (long long i(1); i <= n; i++) s[i] = read();
  for (long long i(1); i <= n; i++) {
    c1 += s[i] / 2;
    c2 += s[i] / 2;
    if (s[i] % 2 == 1) {
      if (i % 2 == 0)
        c2++;
      else
        c1++;
    }
  }
  write(c1 > c2 ? c2 : c1);
  return 0;
}

### Prompt

Please formulate a cpp solution to 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 n, c1, c2, s[300300];
inline long long read() {
  long long w = 1, q = 0;
  char ch = ' ';
  while (ch != '-' && (ch < '0' || ch > '9')) ch = getchar();
  if (ch == '-') w = -1, ch = getchar();
  while (ch >= '0' && ch <= '9') q = q * 10 + ch - '0', ch = getchar();
  return w * q;
}
inline void write(long long x) {
  if (x < 0) {
    x = ~(x - 1);
    putchar('-');
  }
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
signed main() {
  n = read();
  for (long long i(1); i <= n; i++) s[i] = read();
  for (long long i(1); i <= n; i++) {
    c1 += s[i] / 2;
    c2 += s[i] / 2;
    if (s[i] % 2 == 1) {
      if (i % 2 == 0)
        c2++;
      else
        c1++;
    }
  }
  write(c1 > c2 ? c2 : c1);
  return 0;
}

```