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>
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);

        long number = Long.parseLong(scanner.nextLine());
        String[] test = scanner.nextLine().split(" ");

        long[] answer = {0, 0};
        
        for (int i = 0; i < number; i++) {
            answer[0] += Integer.valueOf(test[i]) / 2;
            answer[1] += Integer.valueOf(test[i]) / 2;
            if (Integer.valueOf(test[i]) % 2 != 0) {
                if (i % 2 == 0) {
                    answer[0]++;
                } else {
                    answer[1]++;
                }
            }
        }

//        for (int i = 0; i < number; i++) {
//            if (Integer.valueOf(test[i]) % 2 == 0) {
//                answer[0] += Integer.valueOf(test[i]) / 2;
//                answer[1] += Integer.valueOf(test[i]) / 2;
//            } else if (Integer.valueOf(test[i]) % 2 != 0) {
//                if (i % 2 == 0) {
//                    answer[0] += Integer.valueOf(test[i]) / 2 + 1;
//                    answer[1] += Integer.valueOf(test[i]) / 2;
//                } else {
//                    answer[1] += Integer.valueOf(test[i]) / 2 + 1;
//                    answer[0] += Integer.valueOf(test[i]) / 2;
//                }
//            }
//        }



        System.out.println(Math.min(answer[0], answer[1]));
    }
}
