Today, Wet Shark is given n integers. Using any of these integers no more than once, Wet Shark wants to get maximum possible even (divisible by 2) sum. Please, calculate this value for Wet Shark. 

Note, that if Wet Shark uses no integers from the n integers, the sum is an even integer 0.

Input

The first line of the input contains one integer, n (1 ≤ n ≤ 100 000). The next line contains n space separated integers given to Wet Shark. Each of these integers is in range from 1 to 109, inclusive. 

Output

Print the maximum possible even sum that can be obtained if we use some of the given integers. 

Examples

Input

3
1 2 3


Output

6

Input

5
999999999 999999999 999999999 999999999 999999999


Output

3999999996

Note

In the first sample, we can simply take all three integers for a total sum of 6.

In the second sample Wet Shark should take any four out of five integers 999 999 999.
import java.io.*;
import java.text.DecimalFormat;
import java.util.*;

public class CF_621A {
	public static void main(String[] args) {
		new CF_621A().run();
	}
	public void run(){
		FastScanner sc = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		
		int n = sc.nextInt();
		ArrayList<Integer> odd = new ArrayList<>();
		long sum = 0;
		for(int i = 0;i<n;i++){
			int tmp = sc.nextInt();
			if((tmp&1)!=0)
				odd.add(tmp);
			sum+=tmp;
		}
		if((sum&1)==0)
			out.println(sum);
		else{
			Collections.sort(odd);
			out.println(sum-odd.get(0));
		}
		out.close();
	}
}

class FastScanner {
	public BufferedReader br;
	public StringTokenizer st;

	public FastScanner(InputStream in) {
		br = new BufferedReader(new InputStreamReader(in));
		st = null;
	}

	public boolean hasMoreTokens() {
		while (st == null || !st.hasMoreTokens()) {
			String line = null;
			try {
				line = br.readLine();
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (line == null)
				return false;
			st = new StringTokenizer(line);
		}
		return true;
	}

	public String next() {
		if (!hasMoreTokens())
			return null;
		return st.nextToken();
	}

	public char nextChar(){
		try {
			return (char) br.read();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return 0;
	}
	public String nextLine(){
		try {
			return br.readLine();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return null;
	}
	public int nextInt() {
		return Integer.parseInt(next());
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	//读一个int类型的数组
	public int[] readIntArray(int n) {
		int[] array = new int[n];
		for (int i = 0; i < n; i++) {
			array[i] = nextInt();
		}
		return array;
	}

	//读一个long数组
	public long[] readLongArray(int n) {
		long[] array = new long[n];
		for (int i = 0; i < n; i++) {
			array[i] = nextLong();
		}
		return array;
	}

	//读一个字符串数组
	public String[] readStringArray(int n) {
		String[] a = new String[n];
		for (int i = 0; i < n; i++) {
			a[i] = next();
		}
		return a;
	}

	//打乱一个数组，便于排序
	public void shuffleIntArray(int[] a) {
		Random rand = new Random();
		for (int i = 0; i < a.length; i++) {
			int x = rand.nextInt(i + 1); // 0~i
			int t = a[i];
			a[i] = a[x];
			a[x] = t;
		}
	}

	// 只适合比较小的数，太大的数会越界
	public long factorial(long n) {
		if (n == 0)
			return 1L;
		long res = 1;
		for (long i = 2; i <= n; i++) {
			res *= i;
		}
		return res;
	}

	// 只适合比较小的数，太大的数会越界
	public long combination(long mom, long son) {
		return factorial(mom) / (factorial(son) * factorial(mom - son));
	}

	//求最小公倍数
	public long gcd(long a, long b) {
		return b == 0 ? a : gcd(b, a % b);
	}

	//求最大公约数
	public long lcm(long a, long b) {
		return a * b / (gcd(a, b));
	}

	//格式化double类型的数
	public String format(String pattern, double d) {
		DecimalFormat df = new DecimalFormat(pattern);
		return df.format(d);
	}

	//判断一个数是否是2的指数
	public boolean isPowerOfTwo(long v) {
		return (v & (v - 1)) == 0;
	}
	
	public void outputIntArray(PrintWriter out, int[] a){
		for(int i = 0;i<a.length-1;i++)
			out.print(a[i] + " ");
		out.println(a[a.length-1]);
	}
}
