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 power(long long x, long long y);
long long gcd(long long a, long long b);
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (long long i = 0; i < n; i++) cin >> arr[i];
    vector<long long> temp;
    long long ans = 0;
    for (long long i = 0; i < n; i += 2) ans += arr[i];
    for (long long i = 0; i < n; i += 2) {
      if (i + 1 < n) {
        temp.push_back(arr[i + 1] - arr[i]);
      }
    }
    long long maxsofar = 0, maxending = 0;
    long long start = -1, end = -1;
    long long currstart = 0;
    for (long long i = 0; i < temp.size(); i++) {
      maxending += temp[i];
      if (maxsofar < maxending) {
        maxsofar = maxending;
        start = currstart;
        end = i;
      }
      if (maxending < 0) {
        maxending = 0;
        currstart = i + 1;
      }
    }
    if (start != -1 && end != -1) {
      long long sum = 0;
      for (long long i = 0; i < start * 2; i += 2) sum += arr[i];
      for (long long i = start * 2 + 1; i <= end * 2 + 1; i += 2) sum += arr[i];
      for (long long i = end * 2 + 2; i < n; i += 2) sum += arr[i];
      if (sum > ans) ans = sum;
    }
    temp.clear();
    for (long long i = 1; i < n; i += 2) {
      if (i + 1 < n) {
        temp.push_back(arr[i] - arr[i + 1]);
      }
    }
    maxsofar = 0, maxending = 0;
    start = -1, end = -1;
    currstart = 0;
    for (long long i = 0; i < temp.size(); i++) {
      maxending += temp[i];
      if (maxsofar < maxending) {
        maxsofar = maxending;
        start = currstart;
        end = i;
      }
      if (maxending < 0) {
        maxending = 0;
        currstart = i + 1;
      }
    }
    if (start != -1 && end != -1) {
      long long sum = 0;
      for (long long i = 0; i < start * 2 + 1; i += 2) sum += arr[i];
      for (long long i = start * 2 + 1; i <= end * 2 + 1; i += 2) sum += arr[i];
      for (long long i = end * 2 + 4; i < n; i += 2) sum += arr[i];
      if (sum > ans) ans = sum;
    }
    cout << ans << "\n";
  }
  return 0;
}
long long power(long long x, long long y) {
  long long res = 1;
  x = x % 1000000007;
  while (y > 0) {
    if (y & 1) res = (res * x) % 1000000007;
    y = y >> 1;
    x = (x * x) % 1000000007;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
