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>
int t, n, a[1500000];
long long ans, ans1, f[1500000], g[1500000];
char s[1500000];
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    ans = ans1 = 0;
    for (int i = 0; i < n; i++) {
      scanf("%d", &a[i]);
      if (i % 2 == 0) ans += a[i], a[i] = -a[i];
    }
    f[0] = a[0];
    g[0] = 0;
    for (int i = 1; i < n; i++) {
      f[i] = std::max(1ll * a[i], g[i - 1] + a[i]);
      g[i] = f[i - 1] + a[i];
      ans1 = std::max(ans1, g[i]);
    }
    printf("%lld\n", ans + ans1);
  }
}
