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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long mxn = 3 * 1e5 + 5, inf = 1e9 + 7, mod = 1e9 + 7;
const long long infLL = 1e18 + 5;
const double EPS = 1e-9, PI = acos(-1);
template <typename T>
void print(T begin, T end) {
  for (T i = begin; i != end; i++) cout << *i << ' ';
}
void sread(string &s) {
  cin >> s;
  s = '_' + s;
}
void sprint(string &s) { cout << s.substr(1, s.size()); }
long long n, m, Q, k, a, b, c, sum, ans, MAXans, MINans, sc, pos, val, l, r,
    acc[2][mxn], racc[mxn], ar[mxn];
vector<long long> v, g[mxn];
bitset<mxn> flag;
bool sol;
string s;
long long f(long long x, long long xend) {
  return (x - 1 == 0 ? 0 : acc[0][x - 1]) +
         (acc[1][xend] - (x - 1 == 0 ? 0 : acc[1][x - 1])) +
         (sc - acc[0][xend - 1]);
}
long long f2(long long x, long long xend) {
  return acc[0][x - 1] + (acc[1][xend] - acc[1][x - 1]) + (sc - acc[0][xend]);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long TC;
  cin >> TC;
  while (TC--) {
    cin >> n;
    for (long long i = 0; i < n; i++) {
      cin >> ar[i];
      acc[0][i] = (i == 0 ? 0 : acc[0][i - 1]) + (ar[i] * (i % 2 == 0));
      acc[1][i] = (i == 0 ? 0 : acc[1][i - 1]) + (ar[i] * (i % 2));
    }
    sc = acc[0][n - 1];
    ans = max(acc[0][n - 1], acc[1][n - 1]);
    long long x = 0;
    for (long long i = 1; i < n; i += 2) {
      ans = max(ans, f(x, i));
      if (f(x, i) < f(i - 1, i)) {
        x = i - 1;
      }
      ans = max(ans, f(x, i));
    }
    x = 1;
    for (long long i = 2; i < n; i += 2) {
      ans = max(ans, f2(x, i));
      if (f2(x, i) < f2(i - 1, i)) {
        x = i - 1;
      }
      ans = max(ans, f2(x, i));
    }
    cout << ans << '\n';
    cout << '\n';
  }
  return 0;
}
