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 maxn = 2e5;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (auto &x : a) cin >> x;
    long long curmax = 0;
    long long cur = 0;
    int l, r;
    l = r = 0;
    int lmax = -1, rmax = -1;
    for (int i = 0; i < n - 1; i += 2) {
      int d = a[i + 1] - a[i];
      cur += d;
      if (cur < 0) {
        cur = 0;
        l = i + 2;
        r = i + 2;
        continue;
      }
      r = i + 1;
      if (cur > curmax) {
        lmax = l;
        rmax = r;
        curmax = cur;
      }
    }
    l = 1, r = 1;
    cur = 0;
    for (int i = 1; i < n - 1; i += 2) {
      int d = a[i] - a[i + 1];
      cur += d;
      if (cur < 0) {
        cur = 0;
        l = i + 2;
        r = i + 2;
        continue;
      }
      r = i + 1;
      if (cur > curmax) {
        lmax = l;
        rmax = r;
        curmax = cur;
      }
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      if (lmax <= i && i <= rmax) {
        if (i & 1) ans += a[i];
      } else if (i % 2 == 0) {
        ans += a[i];
      }
    }
    cout << ans << endl;
  }
}
