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 maxSubArraySum(vector<long long> &v) {
  long long sum = 0;
  long long mx = 0;
  for (int i = 0; i < (int)v.size(); i++) {
    sum += v[i];
    sum = max(0LL, sum);
    mx = max(mx, sum);
  }
  return mx;
}
void syntex_error() {
  long long i, j, k, l, r, n, m, a, b, c, x, y, z, sum = 0LL;
  vector<long long> v, one, two;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> x;
    v.push_back(x);
    if (i % 2 == 0) sum += x;
  }
  for (i = 0; i < n; i += 2) {
    if (i > 0) {
      one.push_back(v[i - 1] - v[i]);
    }
    if (i + 1 < n) {
      two.push_back(v[i + 1] - v[i]);
    }
  }
  x = maxSubArraySum(one);
  y = maxSubArraySum(two);
  sum += max(x, y);
  cout << sum << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long TC = 1LL;
  cin >> TC;
  while (TC--) {
    syntex_error();
  }
  return 0;
}
