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.math.*;
import java.io.*;
import java.util.*;
public class Main
{
	StringTokenizer st;
	BufferedReader in;
	PrintWriter ob;
	public static void main(String args[])throws IOException {
		new Main().run();
	}
	void run()throws IOException   {
		
		//in=new BufferedReader(new FileReader("input.txt"));
		//ob=new PrintWriter("output.txt");
		in=new BufferedReader(new InputStreamReader(System.in));
		ob=new PrintWriter(System.out);
		solve();
		ob.flush();
	}
	void solve()throws IOException {
		read();
		int n=ni();
		int a[]=new int[n];
		int b[]=new int[n];
		BigInteger ans=new BigInteger("0");
		read();
		int k=-1;
		for(int i=0;i<n;i++)
		{
			a[i]=ni();
			if(a[i]%2==0)
			ans=	ans.add(new BigInteger(""+a[i]));
			else
				b[++k]=a[i];
		}
	//	System.out.println(ans+" " +k);
		Arrays.sort(b,0,k+1);
	//	System.out.println(k+" " +b[0]+b[1]);
		for(int j=k;j>=0;j-=2)
		{
			if(j-1>=0)
			ans=ans.add(new BigInteger(""+(b[j]+b[j-1])));
		}
		ob.println(ans);
		ob.flush();
	}
	void read()throws IOException  {
		st=new StringTokenizer(in.readLine());
	}
	int ni(){
		return Integer.parseInt(st.nextToken());
	}
	long nl(){
		return Long.parseLong(st.nextToken());
	}
	double nd(){
		return Double.parseDouble(st.nextToken());
	}
	String ns(){
		return st.nextToken();
	}
	char nc(){
		return st.nextToken().charAt(0);
	}
	int[] nia(int n)throws IOException {
		int a[]=new int[n];
		read();
		for(int i=0;i<n;i++){
			a[i]=Integer.parseInt(st.nextToken());
		}
		return a;
	}
	long[] nla(int n)throws IOException {
		long a[]=new long[n];
		read();
		for(int i=0;i<n;i++){
			a[i]=Long.parseLong(st.nextToken());
		}
		return a;
	}
}
