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#include <bits/stdc++.h>
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using namespace std;
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const int N = 1e5 + 10;
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int T; //T组数据
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int n; //每一组数据的个数n
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int a[N]; //每个商店的金钱数量
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/**
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f[i][0] 代表的是不偷第i家店铺能得到的最多现金数量；
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f[i][1] 代表的是偷第i家店铺能得到的最多现金数量。
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*/
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int f[N][2];
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/**
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状态机 O(n)
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把一个过程用一种确定的状态描述了出来
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如 f[i][0] 表示没有偷第 i 个商店, f[i][1] 表示偷了第 i 个商店
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则 f[i][0] 的入边（即过程）有两条 1. 偷了第 i - 1 个商店， 2. 没偷第 i - 1 个商店
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而 f[i][1] 的入边仅有一条，即 没偷第 i - 1 个商店。
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*/
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//状态机解法
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int main() {
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scanf("%d", &T);
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while (T--) {
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scanf("%d", &n);
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memset(f, 0, sizeof f);
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for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
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//初始化
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f[1][0] = 0, f[1][1] = a[1];
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//逐个把商店加入
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for (int i = 2; i <= n; i++) {
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//不偷i号商店,获利取原来前面i-1号商店决策完的最大值
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f[i][0] = max(f[i - 1][0], f[i - 1][1]);
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//偷i号商店，获利取不偷i-1号商店的决策值，再加上当前商店的金额
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f[i][1] = f[i - 1][0] + a[i];
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}
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//最终的结果二选一
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printf("%d\n", max(f[n][0], f[n][1]));
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}
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return 0;
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}
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