#include <bits/stdc++.h>
int temparr1[100001];
int temparr2[100001];
long long maxsum(int n, int *arr) {
  long long int ans = 0;
  long long int temp = 0;
  for (int i = 0; i < n; ++i) {
    temp = temp + arr[i];
    if (temp < 0) {
      temp = 0;
    }
    if (ans < temp) {
      ans = temp;
    }
  }
  return ans;
}
long long int maxevensum(int n, int *arr) {
  int half = n / 2;
  int halfn = (n - 1) / 2;
  long long int sum = 0;
  long long int t1, t2;
  for (int i = 0; i < n / 2; i++) {
    sum = sum + arr[2 * i];
    temparr1[i] = arr[2 * i + 1] - arr[2 * i];
    if (i < (n - 1) / 2) {
      temparr2[i] = arr[2 * i + 1] - arr[2 * i + 2];
    }
  }
  t1 = maxsum(half, temparr1);
  t2 = maxsum(halfn, temparr2);
  if (n % 2 == 1) {
    sum = sum + arr[n - 1];
  } else {
    sum = sum;
  }
  if (t1 > t2) {
    return sum + t1;
  } else {
    return sum + t2;
  }
}
int main() {
  int t, n, arr[200001];
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &arr[i]);
    }
    printf("%lld\n", maxevensum(n, arr));
  }
}

### Prompt

Please provide a Cpp coded solution 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 temparr1[100001];
int temparr2[100001];
long long maxsum(int n, int *arr) {
  long long int ans = 0;
  long long int temp = 0;
  for (int i = 0; i < n; ++i) {
    temp = temp + arr[i];
    if (temp < 0) {
      temp = 0;
    }
    if (ans < temp) {
      ans = temp;
    }
  }
  return ans;
}
long long int maxevensum(int n, int *arr) {
  int half = n / 2;
  int halfn = (n - 1) / 2;
  long long int sum = 0;
  long long int t1, t2;
  for (int i = 0; i < n / 2; i++) {
    sum = sum + arr[2 * i];
    temparr1[i] = arr[2 * i + 1] - arr[2 * i];
    if (i < (n - 1) / 2) {
      temparr2[i] = arr[2 * i + 1] - arr[2 * i + 2];
    }
  }
  t1 = maxsum(half, temparr1);
  t2 = maxsum(halfn, temparr2);
  if (n % 2 == 1) {
    sum = sum + arr[n - 1];
  } else {
    sum = sum;
  }
  if (t1 > t2) {
    return sum + t1;
  } else {
    return sum + t2;
  }
}
int main() {
  int t, n, arr[200001];
  scanf("%d", &t);
  for (int i = 0; i < t; ++i) {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &arr[i]);
    }
    printf("%lld\n", maxevensum(n, arr));
  }
}

```