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;
const int N = 2e5 + 5;
long long a[N], dp[N][3][2], n;
long long solve(int i, int stage, int one) {
  if (i == n) return (stage == 1) * -1e15;
  long long &ret = dp[i][stage][one];
  if (~ret) return ret;
  ret = 0;
  if (stage == 0) {
    ret = !(i & 1) * a[i] + solve(i + 1, stage, one);
    ret = max(ret, (i & 1) * a[i] + solve(i + 1, 1, i % 2));
  } else if (stage == 1) {
    ret = (i & 1) * a[i] + solve(i + 1, stage, one);
    if (one != (i & 1)) ret = max(ret, (i & 1) * a[i] + solve(i + 1, 2, i % 2));
  } else {
    ret = !(i & 1) * a[i] + solve(i + 1, stage, one);
  }
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 3; j++)
        for (int k = 0; k < 2; k++) dp[i][j][k] = -1;
    cout << solve(0, 0, 0) << "\n";
  }
  return 0;
}
