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 long long maxn = 3e5 + 5;
long long a[maxn];
long long dp1[maxn];
long long dp2[maxn];
void mian() {
  long long n;
  cin >> n;
  dp2[0] = LLONG_MIN / 3ll - 2333ll;
  long long sum = 0;
  for (long long i = 1; i <= n; ++i) {
    cin >> a[i];
    if (i % 2) sum += a[i];
    dp1[i] = dp2[i] = LLONG_MIN / 3ll - 2333ll;
  }
  for (long long i = 3; i <= n; i += 2) {
    long long cha = a[i - 1] - a[i];
    if (dp1[i - 2] <= 0) {
      dp1[i] = cha;
    } else {
      dp1[i] = dp1[i - 2];
      dp1[i] += cha;
    }
  }
  for (long long i = 2; i <= n; i += 2) {
    long long cha = a[i] - a[i - 1];
    if (dp2[i - 2] <= 0) {
      dp2[i] = cha;
    } else {
      dp2[i] = dp2[i - 2];
      dp2[i] += cha;
    }
  }
  long long maxn = 0;
  for (long long i = 1; i <= n; ++i) {
    maxn = max(maxn, dp1[i]);
    maxn = max(maxn, dp2[i]);
  }
  cout << sum + maxn << endl;
}
int main() {
  ios::sync_with_stdio(0);
  long long t;
  cin >> t;
  string str;
  for (long long i = 0; i < t; ++i) {
    mian();
  }
  return 0;
}
