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;
int gcd(int i, int j) {
  if (j == 0)
    return i;
  else
    return gcd(j, i % j);
}
template <typename T>
inline T getint() {
  T val = 0;
  char c;
  bool neg = false;
  while ((c = getchar()) && !(c >= '0' && c <= '9')) {
    neg |= c == '-';
  }
  do {
    val = (val * 10) + c - '0';
  } while ((c = getchar()) && (c >= '0' && c <= '9'));
  return val * (neg ? -1 : 1);
}
const long long INF = 10000000;
const int Y = 200100;
const long long m = 1000000007;
long long a[Y];
int32_t main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long s = 0;
    vector<long long> a(n);
    set<long long> b1 = {0};
    set<long long> b2 = {0};
    long long w1 = 0, w2 = 0;
    long long df = 0;
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      if (i % 2 == 0) {
        s += a[i];
        if (i != 0) {
          w2 += a[i - 1] - a[i];
          b2.insert(w2);
          long long t2 = w2 - (*b2.begin());
          df = max(df, t2);
        }
      } else {
        w1 += a[i] - a[i - 1];
        b1.insert(w1);
        long long t1 = w1 - (*b1.begin());
        df = max(df, t1);
      }
    }
    cout << s + df << endl;
  }
  return 0;
}
