#include <bits/stdc++.h>
int main() {
  long long rows, cols;
  scanf("%lld", &rows);
  while (rows--) {
    long long sub_p = 0, sub_unp = 0, max_p = 0, max_unp = 0, sum = 0;
    long long num[2] = {0, 0};
    scanf("%lld", &cols);
    scanf("%lld", &num[1]);
    for (; cols > 0; cols--) {
      num[0] = num[1];
      if (cols > 1) scanf("%d", &num[1]);
      ;
      sub_p = num[1] - num[0] + (sub_p > 0 ? sub_p : 0);
      max_p = (sub_p > max_p ? sub_p : max_p);
      sum += num[0];
      cols--;
      num[0] = num[1];
      if (cols > 1) scanf("%d", &num[1]);
      ;
      sub_unp = num[0] - num[1] + (sub_unp > 0 ? sub_unp : 0);
      max_unp = (sub_unp > max_unp ? sub_unp : max_unp);
    }
    printf("%lld\n", sum + (max_p > max_unp ? max_p : max_unp));
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
int main() {
  long long rows, cols;
  scanf("%lld", &rows);
  while (rows--) {
    long long sub_p = 0, sub_unp = 0, max_p = 0, max_unp = 0, sum = 0;
    long long num[2] = {0, 0};
    scanf("%lld", &cols);
    scanf("%lld", &num[1]);
    for (; cols > 0; cols--) {
      num[0] = num[1];
      if (cols > 1) scanf("%d", &num[1]);
      ;
      sub_p = num[1] - num[0] + (sub_p > 0 ? sub_p : 0);
      max_p = (sub_p > max_p ? sub_p : max_p);
      sum += num[0];
      cols--;
      num[0] = num[1];
      if (cols > 1) scanf("%d", &num[1]);
      ;
      sub_unp = num[0] - num[1] + (sub_unp > 0 ? sub_unp : 0);
      max_unp = (sub_unp > max_unp ? sub_unp : max_unp);
    }
    printf("%lld\n", sum + (max_p > max_unp ? max_p : max_unp));
  }
  return 0;
}

```