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;
const int MAXN = 2e5 + 100;
const long long INF = 1e18;
int n;
long long a[MAXN];
long long dp[MAXN][3][2];
long long go(int pos, int open, int len) {
  if (pos == n) {
    if (open == 1 && len == 0) return -INF;
    return 0;
  }
  if (dp[pos][open][len] != -1) return dp[pos][open][len];
  int nlen = 0;
  if (open == 1) nlen = 1 - len;
  long long op1 = go(pos + 1, open, nlen);
  if ((open == 0 || open == 2) && pos % 2 == 0) op1 += a[pos];
  if (open == 1 && pos % 2 == 1) op1 += a[pos];
  int nopen = -1;
  long long op2 = 0;
  if (open == 0) nopen = 1;
  if (open == 1 && len == 1) {
    nopen = 2;
    nlen = 0;
  }
  if (nopen != -1) {
    op2 = go(pos + 1, nopen, nlen);
    if ((nopen == 0 || nopen == 2) && pos % 2 == 0) op2 += a[pos];
    if (nopen == 1 && pos % 2 == 1) op2 += a[pos];
  }
  return dp[pos][open][len] = max(op1, op2);
}
long long solve() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n + 1; i++) {
    for (int j = 0; j < 3; j++) {
      for (int k = 0; k < 2; k++) {
        dp[i][j][k] = -1;
      }
    }
  }
  long long res = go(0, 0, 0);
  return res;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T;
  cin >> T;
  while (T--) {
    cout << solve() << endl;
  }
  return 0;
}
