CF1915E.Romantic Glasses

Iulia has $n$ glasses arranged in a line. The $i$-th glass has $a_i$ units of juice in it. Iulia drinks only from odd-numbered glasses, while her date drinks only from even-numbered glasses.

To impress her date, Iulia wants to find a contiguous subarray of these glasses such that both Iulia and her date will have the same amount of juice in total if only the glasses in this subarray are considered. Please help her to do that.

More formally, find out if there exists two indices $l$, $r$ such that $1 \leq l \leq r \leq n$, and $a_l + a_{l + 2} + a_{l + 4} + \dots + a_{r} = a_{l + 1} + a_{l + 3} + \dots + a_{r-1}$ if $l$ and $r$ have the same parity and $a_l + a_{l + 2} + a_{l + 4} + \dots + a_{r - 1} = a_{l + 1} + a_{l + 3} + \dots + a_{r}$ otherwise.

尤莉娅有 $n$ 个杯子排成一行。第 $i$ 个杯子中有 $a_i$ 单位果汁。尤莉娅只从编号为奇数的杯子中饮用，而她的约会对象只从编号为偶数的杯子中饮用。

为了给约会对象留下深刻印象，尤莉娅希望找出一个连续子数组（即一段连续的杯子），使得仅考虑该子数组中的杯子时，她和约会对象所饮果汁总量相等。请帮她解决这个问题。

更形式化地说：判断是否存在两个下标 $l$、$r$，满足 $1 \leq l \leq r \leq n$，且当 $l$ 与 $r$ 同为奇数或同为偶数时，有

$$a_l + a_{l + 2} + a_{l + 4} + \dots + a_{r} = a_{l + 1} + a_{l + 3} + \dots + a_{r-1},$$

否则（即 $l$ 与 $r$ 奇偶性不同）有

$$a_l + a_{l + 2} + a_{l + 4} + \dots + a_{r - 1} = a_{l + 1} + a_{l + 3} + \dots + a_{r}.$$

The first line contains a single integer $t$ ($1 \leq t \leq 10^4$) — the number of test cases.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 2 \cdot 10^5$) — the total number of glasses.

The second line of each test case contains $n$ integers $a_1, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the amount of juice in each glass.

The sum of $n$ over all test cases does not exceed $2 \cdot 10^5$.

第一行包含一个整数 $t$（$1 \leq t \leq 10^4$）—— 测试用例的数量。

每个测试用例的第一行包含一个整数 $n$（$1 \leq n \leq 2 \cdot 10^5$）—— 玻璃杯的总数。

每个测试用例的第二行包含 $n$ 个整数 $a_1, \ldots, a_n$（$1 \leq a_i \leq 10^9$）—— 每个玻璃杯中果汁的量。

所有测试用例的 $n$ 之和不超过 $2 \cdot 10^5$。

For each test case, output "YES" if there exists a subarray satisfying the condition, and "NO" otherwise.

You can output the answer in any case (for example, the strings "yEs", "yes", "Yes" and "YES" will be recognized as a positive answer).

对于每个测试用例，如果存在满足条件的子数组，则输出 “YES”，否则输出 “NO”。

您可以以任意大小写形式输出答案（例如，字符串 “yEs”、“yes”、“Yes” 和 “YES” 均会被识别为肯定回答）。

In the first test case, Iulia can pick $l=1$ and $r=3$. Then she drinks $a_1+a_3=1+2=3$ units and her date drinks $a_2=3$ units of juice.

In the second test case, Iulia can pick $l=2$ and $r=5$. Then she drinks $a_3+a_5=1+1=2$ units and her date drinks $a_2+a_4=1+1=2$ units of juice.

In the third test case no such contiguous subarray works.

In the fourth test case, Iulia can pick $l=2$ and $r=8$. Then she drinks $a_3+a_5+a_7=11+1+1=13$ units and her date drinks $a_2+a_4+a_6+a_8=2+4+5+2=13$ units of juice.

在第一个测试用例中，Iulia 可以选择 $l=1$ 和 $r=3$。此时她喝掉 $a_1+a_3=1+2=3$ 单位果汁，而她的约会对象喝掉 $a_2=3$ 单位果汁。

在第二个测试用例中，Iulia 可以选择 $l=2$ 和 $r=5$。此时她喝掉 $a_3+a_5=1+1=2$ 单位果汁，而她的约会对象喝掉 $a_2+a_4=1+1=2$ 单位果汁。

在第三个测试用例中，不存在满足条件的连续子数组。

在第四个测试用例中，Iulia 可以选择 $l=2$ 和 $r=8$。此时她喝掉 $a_3+a_5+a_7=11+1+1=13$ 单位果汁，而她的约会对象喝掉 $a_2+a_4+a_6+a_8=2+4+5+2=13$ 单位果汁。