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;
bool comp(int i, int j) { return (i < j); }
long long err(long long x, long long y, long long z) {
  return (((x - y) / 2) * ((x - y) / 2) + ((x - z) / 2) * ((x - z) / 2) +
          ((y - z) / 2) * ((y - z) / 2));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  double f, e1, e2, pi;
  int a[200005], b[200005], c[200005];
  long long dd1, dd0, bb, cc, tt, dd, i, j, maxx, a0, rrr, minn, ans_l, ans_r,
      min_sum, min_pos, sum, sum1, sum2, sum3, sum4, sum5, k1, k2, kk1, kk2;
  long long y, x, u, h, a1, aa, k, n, t, ch,
      zz = 0, kl, odn = 1, m, m1, m2, m0, m3, m4, xl, xr, yl, yr, i1, i2, a2,
      i3, nn, x1, x2, y1, y2, z11, z1, z2, z3, z4;
  int l, r;
  set<long long> nk;
  l = 0;
  h = 0;
  pi = atan(1) * 4;
  ch = 0;
  bb = 0;
  aa = 0;
  a0 = 0;
  zz = 20000000000;
  k = 0;
  r = 0;
  t = 0;
  j = 0;
  zz = 0;
  k1 = 0;
  k2 = 0;
  kk1 = 0;
  kk2 = 0;
  m1 = 0;
  m2 = 0;
  m3 = 0;
  dd0 = 0;
  dd1 = 0;
  m4 = 0;
  aa = 0;
  sum = 0;
  vector<vector<int> > g;
  string s, s1, s2;
  cin >> kk2;
  for (kk1 = 0; kk1 < kk2; kk1++) {
    sum = 0;
    cin >> n;
    k = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      if (i % 2 == 0) k += a[i];
    }
    m1 = 0;
    m2 = 0;
    for (i = 0; i < n - 1; i += 2) {
      b[m1] = a[i + 1] - a[i];
      m1++;
    }
    for (i = 1; i < n - 1; i += 2) {
      c[m2] = a[i] - a[i + 1];
      m2++;
    }
    sum1 = b[0];
    a0 = 0;
    sum2 = 0;
    for (int i = 0; i < m1; i++) {
      sum2 += b[i];
      if (sum2 - a0 > sum1) sum1 = sum2 - a0;
      if (a0 > sum2) a0 = sum2;
    }
    sum4 = c[0];
    a0 = 0;
    sum2 = 0;
    for (int i = 0; i < m2; i++) {
      sum2 += c[i];
      if (sum2 - a0 > sum4) sum4 = sum2 - a0;
      if (a0 > sum2) a0 = sum2;
    }
    if (sum4 > sum1) sum1 = sum4;
    if (sum1 > 0) k += sum1;
    cout << k << endl;
  }
  return 0;
}
