You are given an array a consisting of n integers. Indices of the array start from zero (i. e. the first element is a_0, the second one is a_1, and so on).

You can reverse at most one subarray (continuous subsegment) of this array. Recall that the subarray of a with borders l and r is a[l; r] = a_l, a_{l + 1}, ..., a_{r}.

Your task is to reverse such a subarray that the sum of elements on even positions of the resulting array is maximized (i. e. the sum of elements a_0, a_2, ..., a_{2k} for integer k = ⌊(n-1)/(2)⌋ should be maximum possible).

You have to answer t independent test cases.

Input

The first line of the input contains one integer t (1 ≤ t ≤ 2 ⋅ 10^4) — the number of test cases. Then t test cases follow.

The first line of the test case contains one integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the length of a. The second line of the test case contains n integers a_0, a_1, ..., a_{n-1} (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

It is guaranteed that the sum of n does not exceed 2 ⋅ 10^5 (∑ n ≤ 2 ⋅ 10^5).

Output

For each test case, print the answer on the separate line — the maximum possible sum of elements on even positions after reversing at most one subarray (continuous subsegment) of a.

Example

Input


4
8
1 7 3 4 7 6 2 9
5
1 2 1 2 1
10
7 8 4 5 7 6 8 9 7 3
4
3 1 2 1


Output


26
5
37
5
//package Problemset.Tag.DynamicProgramming;
import java.io.*;

public class MaximumSumAtEvenPositionsDP {
    public static void main(String[] args) throws Exception{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(br.readLine());
        String[] str;
        while(t-->0) {
            int n = Integer.parseInt(br.readLine());
            str = br.readLine().split(" ");
            long[] arr = new long[n];
            for(int i=0; i<n; i++) arr[i] = Long.parseLong(str[i]);
            long[][] dp = new long[n+1][3];
            for(int i=0; i<n; i++) {
                dp[i+1][0] = Math.max(dp[i+1][0], dp[i][0]+((i%2==0)?arr[i]:0));
                if(i+2<=n)
                    dp[i+2][1] = Math.max(dp[i+2][1], Math.max(dp[i][0], dp[i][1])+((i%2==0)?arr[i+1]:arr[i]));
                dp[i+1][2] = Math.max(dp[i+1][2], Math.max(Math.max(dp[i][1], dp[i][2]), dp[i][1]) +((i%2==0)?arr[i]:0));
            }
       
            bw.write(Math.max(Math.max(dp[n][0], dp[n][1]), dp[n][2])+"\n");
        }
        bw.close();
        br.close();
    }
}