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 fx[] = {+1, -1, +0, +0};
const int fy[] = {+0, +0, +1, -1};
void err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << endl;
  err(++it, args...);
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
const long long int N = 2e5 + 10;
long long int kadane(vector<long long int> vec, long long int n) {
  long long int mxSoFar = LONG_MIN, mxEndHere = 0;
  long long int lb = 0, ub = 0, s = 0;
  for (long long int i = 0; i < n; i++) {
    mxEndHere += vec[i];
    if (mxSoFar < mxEndHere) {
      mxSoFar = mxEndHere;
      lb = s;
      ub = i;
    }
    if (mxEndHere < 0) {
      mxEndHere = 0;
      s = i + 1;
    }
  }
  return mxSoFar;
}
long long int a[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    vector<long long int> b, c;
    for (long long int i = 1; i < n; i += 2) {
      b.push_back(a[i] - a[i - 1]);
    }
    for (long long int i = 2; i < n; i += 2) {
      c.push_back(a[i - 1] - a[i]);
    }
    long long int sum = 0;
    for (long long int i = 0; i < n; i += 2) {
      sum += a[i];
    }
    long long int cnt =
        sum + max({0LL, kadane(b, b.size()), kadane(c, c.size())});
    cout << cnt << endl;
  }
  return 0;
}
