CF1438B.Valerii Against Everyone

You're given an array $b$ of length $n$. Let's define another array $a$, also of length $n$, for which $a_i = 2^{b_i}$ ($1 \leq i \leq n$).

Valerii says that every two non-intersecting subarrays of $a$ have different sums of elements. You want to determine if he is wrong. More formally, you need to determine if there exist four integers $l_1,r_1,l_2,r_2$ that satisfy the following conditions:

• $1 \leq l_1 \leq r_1 \lt l_2 \leq r_2 \leq n$;
• $a_{l_1}+a_{l_1+1}+\ldots+a_{r_1-1}+a_{r_1} = a_{l_2}+a_{l_2+1}+\ldots+a_{r_2-1}+a_{r_2}$.

If such four integers exist, you will prove Valerii wrong. Do they exist?

An array $c$ is a subarray of an array $d$ if $c$ can be obtained from $d$ by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.

给你一个长度为 $n$ 的数组 $b$。我们定义另一个长度同样为 $n$ 的数组 $a$，其中 $a_i = 2^{b_i}$（$1 \leq i \leq n$）。

Valerii 声称：数组 $a$ 的任意两个互不相交的子数组的元素和均不相等。你想判断他是否说错了。更准确地说，你需要判断是否存在四个整数 $l_1, r_1, l_2, r_2$，满足以下条件：

• $1 \leq l_1 \leq r_1 \lt l_2 \leq r_2 \leq n$；
• $a_{l_1}+a_{l_1+1}+\ldots+a_{r_1-1}+a_{r_1} = a_{l_2}+a_{l_2+1}+\ldots+a_{r_2-1}+a_{r_2}$。

若存在这样的四个整数，则说明 Valerii 的说法是错误的。是否存在这样的四个整数？

数组 $c$ 是数组 $d$ 的子数组，当且仅当 $c$ 可通过从 $d$ 的开头删除若干（可能为零个或全部）元素、并从结尾删除若干（可能为零个或全部）元素而得到。

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 100$). Description of the test cases follows.

The first line of every test case contains a single integer $n$ ($2 \le n \le 1000$).

The second line of every test case contains $n$ integers $b_1,b_2,\ldots,b_n$ ($0 \le b_i \le 10^9$).

每个测试包含多个测试用例。第一行包含测试用例的数量 $t$（$1 \le t \le 100$）。随后是测试用例的描述。

每个测试用例的第一行包含一个整数 $n$（$2 \le n \le 1000$）。

每个测试用例的第二行包含 $n$ 个整数 $b_1,b_2,\ldots,b_n$（$0 \le b_i \le 10^9$）。

For every test case, if there exist two non-intersecting subarrays in $a$ that have the same sum, output YES on a separate line. Otherwise, output NO on a separate line.

Also, note that each letter can be in any case.

对于每个测试用例，如果数组 $a$ 中存在两个不相交的子数组，且它们的和相等，则在单独一行输出 YES；否则，在单独一行输出 NO。

另外，请注意：每个字母可以是任意大小写。

In the first case, $a = [16,8,1,2,4,1]$. Choosing $l_1 = 1$, $r_1 = 1$, $l_2 = 2$ and $r_2 = 6$ works because $16 = (8+1+2+4+1)$.

In the second case, you can verify that there is no way to select to such subarrays.

在第一种情况下，$a = [16,8,1,2,4,1]$。选择 $l_1 = 1$、$r_1 = 1$、$l_2 = 2$ 和 $r_2 = 6$ 是可行的，因为 $16 = (8+1+2+4+1)$。

在第二种情况下，可以验证不存在满足条件的两个子数组。