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 + 2, M = 1e9 + 7;
long long dp[N][3];
int n, a[N];
long long calc(int idx, int rev) {
  if (idx == n) return 0;
  long long &ret = dp[idx][rev];
  if (ret != -1) return ret;
  ret = 0;
  if (!rev) {
    ret = max(ret, ((idx & 1) ? 0 : a[idx]) + calc(idx + 1, 0));
    if (idx + 1 < n) {
      ret = max(ret, ((idx & 1) ? a[idx] : a[idx + 1]) + calc(idx + 2, 1));
    }
  } else if (rev == 1) {
    if (idx + 1 < n) {
      ret = max(ret, ((idx & 1) ? a[idx] : a[idx + 1]) + calc(idx + 2, 1));
    }
    ret = max(ret, ((idx & 1) ? 0 : a[idx]) + calc(idx + 1, 2));
  } else {
    ret = max(ret, ((idx & 1) ? 0 : a[idx]) + calc(idx + 1, 2));
  }
  return ret;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
    }
    for (int i = 0; i < n; ++i) {
      dp[i][0] = -1;
      dp[i][1] = -1;
      dp[i][2] = -1;
    }
    cout << calc(0, 0) << endl;
  }
}
