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;
using ll = long long;
using ld = long double;
using ull = unsigned long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
using vi = vector<int>;
using vll = vector<ll>;
using db = double;
int dx[4] = {-1, 0, 1, 0};
int dy[4] = {0, -1, 0, 1};
ll gcd(ll a, ll b) {
  if (b == 0) {
    return a;
  } else {
    return gcd(b, a % b);
  }
}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
int getInt(char c) { return (int)(c - '0'); }
bool _intervalsIntersect(int a, int b, int c, int d, bool repeat) {
  if ((c >= a) && (c <= b)) {
    return true;
  }
  if ((d >= a) && (d <= b)) {
    return true;
  }
  if (repeat) {
    return _intervalsIntersect(c, d, a, b, false);
  }
  return false;
}
bool intervalsIntersect(int a, int b, int c, int d) {
  return _intervalsIntersect(a, b, c, d, true);
}
string toBin(ull x, int m) {
  string res = "";
  while (x > 0) {
    if (x % 2 == 1) {
      res = "1" + res;
    } else {
      res = "0" + res;
    }
    x /= 2;
  }
  m -= ((int)(res).size());
  for (int k = 0; k < m; k++) {
    res = "0" + res;
  }
  return res;
}
ull sumN(ull x) { return x * (x + 1) / 2; }
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cout.setf(ios::fixed), cout.precision(20);
  int T;
  cin >> T;
  while (T--) {
    ll n;
    cin >> n;
    vector<ll> a(n), sum;
    for (long long k = (0); k < (n); ++k) {
      cin >> a[k];
    }
    if (n == 1) {
      cout << a[0] << endl;
    } else if (n == 2) {
      cout << max(a[0], a[1]) << endl;
    } else {
      for (long long k = (1); k < (n); ++k) {
        if (k % 2 == 1) {
          sum.push_back(a[k] - a[k - 1]);
        } else {
          sum.push_back(a[k - 1] - a[k]);
        }
      }
      ll bestRes = max((ll)0, sum[0]);
      bestRes = max(bestRes, sum[1]);
      ll worst[] = {sum[0], sum[1]};
      for (long long k = (2); k < (((int)(sum).size())); ++k) {
        sum[k] += sum[k - 2];
        bestRes = max(bestRes, sum[k] - worst[k % 2]);
        bestRes = max(bestRes, sum[k]);
        worst[k % 2] = min(worst[k % 2], sum[k]);
      }
      ll res = 0;
      for (long long k = (0); k < (n); ++k) {
        if (k % 2 == 0) {
          res += a[k];
        }
      }
      cout << res + bestRes << endl;
    }
  }
  return 0;
}
