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 Max = 2e5 + 5;
const int Mod = 1e9 + 7;
long long data[Max], ndata[2][Max];
int main() {
  std::ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long sum = 0, osum = 0;
    int cnt1 = 0, cnt2 = 0;
    for (int i = 0; i < n; i++) {
      cin >> data[i];
      if (i & 1) {
        ndata[0][cnt1++] = data[i] - data[i - 1];
      } else {
        sum += data[i];
        if (i) ndata[1][cnt2++] = data[i - 1] - data[i];
      }
    }
    long long minn = 0, per = 0, sub = 0, ans = 0;
    for (int i = 0; i < cnt1; i++) {
      per += ndata[0][i];
      minn = min(minn, per);
      sub = max(sub, per - minn);
    }
    ans = sum + sub;
    per = 0;
    minn = 0;
    for (int i = 0; i < cnt2; i++) {
      per += ndata[1][i];
      minn = min(minn, per);
      sub = max(sub, per - minn);
    }
    ans = max(ans, sum + sub);
    cout << ans << '\n';
  }
}
