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;
long long modI(long long a, long long m);
long long gcd(long long a, long long b);
long long powM(long long x, long long y, long long m);
long long swap(long long a, long long b);
void swap(long long& a, long long& b) {
  long long tp = a;
  a = b;
  b = tp;
}
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powM(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x %= m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans *= x;
      ans %= m;
    }
    r <<= 1;
    x *= x;
    x %= m;
  }
  return ans;
}
long long modI(long long a, long long m) {
  long long m0 = m, y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m;
    a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(0);
  long long tt = 1;
  cin >> tt;
  while (tt--) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    long long sum = 0;
    for (long long i = 0; i < n; i++) {
      if (i % 2 == 0) sum = sum + a[i];
    }
    vector<long long> v;
    for (long long i = 0; i < n; i++) {
      bool ok = 0;
      if (i <= n - 2) ok = 1;
      if (ok) v.push_back(a[i + 1] - a[i]);
      i++;
    }
    long long quant = v.size();
    long long tpmax = 0, tpsum = 0;
    for (long long i = 0; i < quant; i++) {
      tpsum = tpsum + v[i];
      tpmax = max(tpmax, tpsum);
      bool ok = 0;
      if (tpsum < 0) ok = 1;
      if (ok) tpsum = 0;
    }
    vector<long long> vec;
    for (long long i = 1; i < n; i++) {
      if (i <= n - 2) vec.push_back(a[i] - a[i + 1]);
      i++;
    }
    quant = vec.size();
    tpsum = 0;
    for (long long i = 0; i < quant; i++) {
      tpsum = tpsum + vec[i];
      tpmax = max(tpmax, tpsum);
      if (tpsum < 0) tpsum = 0;
    }
    cout << sum + tpmax << endl;
  }
  return 0;
}
