A13958. 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. $1 \leq l_1 \leq r_1 \lt l_2 \leq r_2 \leq n$ ;
2. $a_{l_1}+a_{l_1+1}+\ldots+a_{r_1-1}+a_{r_1} = a_{l_2}+a_{l_2+1}+\ldots+a_{r_2-1}+a_{r_2}$ .

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.

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 $b_1,b_2,\ldots,b_n$ ( $0 \le b_i \le 10^9$ ).

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.

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.