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
import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.*;

public class D {
    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(bf.readLine());
        for (int i = 0; i < t; i++) {
            int n = Integer.parseInt(bf.readLine());
            StringTokenizer st = new StringTokenizer(bf.readLine());
            int[] arr = new int[n];
            for (int j = 0; j < n; j++) {
                arr[j] = Integer.parseInt(st.nextToken());
            }
            // two possibilities:
            // start flip odd, end flip even
            // start flip even, end flip odd
            long best = 0;

            long dp1 = 0;
            long dp2 = 0;
            long dp3 = 0;

            long ndp1 = 0;
            long ndp2 = 0;
            long ndp3 = 0;

            // end even
            for (int j = 0; j < n; j++) {
                if (j % 2 == 0) { // can end
                    ndp1 = dp1 + arr[j];
                    ndp2 = dp2;
                    ndp3 = max(dp3 + arr[j], dp2);
                } else { // can start
                    ndp1 = dp1;
                    ndp2 = max(dp2 + arr[j], dp1 + arr[j]);
                    ndp3 = dp3;
                }
                dp1 = ndp1;
                dp2 = ndp2;
                dp3 = ndp3;
            }
            best = max(best, max(dp1, dp3));

            dp1 = 0;
            dp2 = 0;
            dp3 = 0;

            // start even
            for (int j = 0; j < n; j++) {
                if (j % 2 == 0) { // can start
                    ndp1 = dp1 + arr[j];
                    ndp2 = max(dp2, dp1);
                    ndp3 = dp3 + arr[j];
                } else { // can end
                    ndp1 = dp1;
                    ndp2 = dp2 + arr[j];
                    ndp3 = max(dp3, dp2 + arr[j]);
                }
                dp1 = ndp1;
                dp2 = ndp2;
                dp3 = ndp3;
            }
            best = max(best, max(dp1, dp3));

            out.println(best);
        }
        out.close();
        System.exit(0);
    }

    /*
    public static long doDp(int[] arr, int n, boolean method) {
        long[][] dp = new long[n][3];
        dp[0][0] = arr[0];
        dp[0][1] = 0;
        dp[0][2] = arr[0];
        for (int j = 1; j < n; j++) {
            if (j % 2 == 0) {
                dp[j][0] = dp[j-1][0] + arr[j];
            } else {
                dp[j][0] = dp[j-1][0];
            }

            if (j % 2 == 1) {
                dp[j][1] = dp[j-1][1] + arr[j];
                dp[j][1] = max(dp[j][1], dp[j-1][0] - arr[j-1] + arr[j]);
            } else {
                dp[j][1] = dp[j-1][1];
                dp[j][1] = max(dp[j][1], dp[j-1][0] + arr[j-1]);
            }

            if (j % 2 == 0) {
                dp[j][2] = dp[j-1][2] + arr[j];
            } else {
                dp[j][2] = dp[j-1][2];
            }
        }
        long best = 0;
        for (int j = 0; j < n; j++) {
            for (int k = 0; k < 3; k++) {
                best = max(best, dp[j][k]);
            }
        }
    }*/

    public static long max(long a, long b) {
        return a > b ? a : b;
    }
}