This is the easy version of the problem. The difference between the versions is that in this version, $l=r$. You can hack only if you solved all versions of this problem.
You are given a positive integer $n$ and the first $n$ terms of an infinite binary sequence $a$, which is defined as follows:
- For
$m \gt n$,$a_m = a_1 \oplus a_2 \oplus \ldots \oplus a_{\lfloor \frac{m}{2} \rfloor}$$^{\text{∗}}$.
Your task is to compute the sum of elements in a given range $[l, r]$: $a_l + a_{l + 1} + \ldots + a_r$.
$^{\text{∗}}$$\oplus$ denotes the bitwise XOR operation.
Input
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.
The first line of each test case contains three integers $n$, $l$, and $r$ ($1 \le n \le 2 \cdot 10^5$, $1 \le l=r\le 10^{18}$).
The second line contains $n$ integers $a_1, a_2, \ldots, a_n$ ($\color{red}{a_i \in \{0, 1\}}$) — the first $n$ terms of the sequence $a$.
It is guaranteed that the sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.
Output
For each test case, output a single integer — the sum of elements in the given range.
Note
In the first test case, the sequence $a$ is equal to $$[\underline{\color{red}{1}}, 1, 1, 0, 0, 1, 1, 1, 1, 1, \ldots]$$ where $l = 1$, and $r = 1$. The sum of elements in the range $[1, 1]$ is equal to $$a_1 = 1.$$
In the second test case, the sequence $a$ is equal to $$[\color{red}{1}, \color{red}{0}, \underline{1}, 1, 1, 0, 0, 1, 1, 0, \ldots]$$ where $l = 3$, and $r = 3$. The sum of elements in the range $[3, 3]$ is equal to $$a_3 = 1.$$
样例
9
1 1 1
1
2 3 3
1 0
3 5 5
1 1 1
1 234 234
0
5 1111 1111
1 0 1 0 1
1 1000000000000000000 1000000000000000000
1
10 87 87
0 1 1 1 1 1 1 1 0 0
12 69 69
1 0 0 0 0 1 0 1 0 1 1 0
13 46 46
0 1 0 1 1 1 1 1 1 0 1 1 1
1
1
0
0
1
0
1
0
0
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