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
#include <bits/stdc++.h>
using namespace std;
long long q, n, i, sum, m[2], f[2][200002], kq, a[200002];
int main() {
  cin >> q;
  while (q--) {
    cin >> n;
    sum = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) sum += a[i];
      f[0][i] = f[0][i - 1];
      f[1][i] = f[1][i - 1];
      if (i % 2 == 0)
        f[0][i] += a[i];
      else
        f[1][i] += a[i];
    }
    kq = sum;
    m[1] = 0;
    m[0] = -round(1e9);
    for (i = 0; i < n; i++) {
      int ch = i % 2;
      kq = max(kq, sum + f[1][i - 1] - f[0][i - 1] + m[1 - ch]);
      kq = max(kq, sum + f[1][i - 1] - f[0][i - 1] + m[ch] + a[i] * ch -
                       a[i] * (1 - ch));
      if (i % 2 == 0)
        m[0] = max(m[0], f[0][i] - f[1][i]);
      else
        m[1] = max(m[1], f[0][i] - f[1][i]);
    }
    cout << kq << '\n';
  }
}
