Problem:
Score : 600 points

Problem Statement

Given is an integer sequence A_1, ..., A_N of length N.

We will choose exactly \left\lfloor \frac{N}{2} \right\rfloor elements from this sequence so that no two adjacent elements are chosen.

Find the maximum possible sum of the chosen elements.

Here \lfloor x \rfloor denotes the greatest integer not greater than x.

Constraints

2 \leq N \leq 2\times 10^5

|A_i|\leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

Print the maximum possible sum of the chosen elements.

Sample Input 1

6
1 2 3 4 5 6

Sample Output 1

12

Choosing 2, 4, and 6 makes the sum 12, which is the maximum possible value.

Sample Input 2

5
-1000 -100 -10 0 10

Sample Output 2

0

Choosing -10 and 10 makes the sum 0, which is the maximum possible value.

Sample Input 3

10
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 3

5000000000

Watch out for overflow.

Sample Input 4

27
18 -28 18 28 -45 90 -45 23 -53 60 28 -74 -71 35 -26 -62 49 -77 57 24 -70 -93 69 -99 59 57 -49

Sample Output 4

295

Solution in C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
long long max(long long x, long long y) {
	long long z;
	z = x;
	if (y > x)z = y;
	return z;
}
long long n, w;
long long a[200000];
long long dp[2][2];
long long e[2][2];
int main() {
	scanf("%lld", &n);
	for (long long i = 0; i < n; i++) {
		scanf("%lld", a + i);
	}
	for (long long i = 0; i < n; i++) {
		e[0][0] = dp[0][0];
		e[0][1] = dp[0][1];
		e[1][0] = dp[1][0];
		e[1][1] = dp[1][1];
		if (i % 2 == 0) {
			dp[0][0] = max(e[0][1], e[1][1]);
			dp[1][1] = e[0][1] + a[i];
			dp[1][0] = e[0][0] + a[i];
			dp[0][1] = -1000000000;
			dp[0][1] = -1000000;
		}
		else {
			dp[0][0] = max(e[0][0], e[1][0]);
			dp[0][1] = max(e[0][1], e[1][1]);
			dp[1][1] = e[0][0] + a[i];
			dp[1][0] = -1000000000;
			dp[1][0] *= 1000000;
		}
	}
	if (n % 2) {
		printf("%lld\n", max(dp[0][0], dp[1][0]));
	}
	else printf("%lld\n", max(dp[0][1], dp[1][1]));
}