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 int ANS = 0;
long long int mod = 1000000007;
int gcd(int a, int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
int foo(int n) {
  if (n <= 0) return 1;
  return foo((n * 2) / 3) + foo(n - 2);
}
int go_l(string& s, int l, int r, char c, int napr) {
  int siz = r - l + 1;
  int ln, rn;
  int count = 0;
  if (siz == 1) {
    if (s[l] == c)
      return 0;
    else
      return 1;
  }
  if (napr == 0) {
    for (int i = l; i < l + siz / 2; ++i) {
      if (s[i] != c) count++;
    }
    ln = l + siz / 2;
    rn = r;
    return count + min(go_l(s, ln, rn, c + 1, 0), go_l(s, ln, rn, c + 1, 1));
  }
  if (napr == 1) {
    for (int i = l + siz / 2; i <= r; ++i) {
      if (s[i] != c) count++;
    }
    ln = l;
    rn = l + siz / 2 - 1;
    return count + min(go_l(s, ln, rn, c + 1, 0), go_l(s, ln, rn, c + 1, 1));
  }
}
int main() {
  int t, n;
  cin >> t;
  vector<long long int> a;
  vector<long long int> b;
  vector<long long int> c;
  while (t != 0) {
    --t;
    a.clear();
    b.clear();
    c.clear();
    cin >> n;
    a.resize(n);
    for (int i = 0; i < n; ++i) cin >> a[i];
    long long int sum = 0;
    long long int per = 0;
    for (int i = 0; i < n; i += 2) sum += a[i];
    for (int i = 0; i < n - 1; i += 2) {
      b.push_back(a[i + 1] - a[i]);
    }
    for (int i = 1; i < n - 1; i += 2) {
      c.push_back(a[i] - a[i + 1]);
    }
    long long int tmp = 0;
    for (int i = 0; i < b.size(); ++i) {
      if (b[i] > 0) {
        tmp += b[i];
        if (tmp > per) per = tmp;
        continue;
      }
      if (tmp > per) per = tmp;
      if (tmp + b[i] < 0)
        tmp = 0;
      else
        tmp += b[i];
    }
    tmp = 0;
    for (int i = 0; i < c.size(); ++i) {
      if (c[i] > 0) {
        tmp += c[i];
        if (tmp > per) per = tmp;
        continue;
      }
      if (tmp > per) per = tmp;
      if (tmp + c[i] < 0)
        tmp = 0;
      else
        tmp += c[i];
    }
    cout << sum + per << endl;
  }
  return 0;
}
