Di-visible Confusion
Contest: Codeforces Round 752 (Div. 1)
Difficulty rating: 1300
Tags: constructive algorithms, math, number theory

YouKn0wWho has an integer sequence $$$a_1, a_2, \ldots, a_n$$$. He will perform the following operation until the sequence becomes empty: select an index $$$i$$$ such that $$$1 \le i \le |a|$$$ and $$$a_i$$$ is not divisible by $$$(i + 1)$$$, and erase this element from the sequence. Here $$$|a|$$$ is the length of sequence $$$a$$$ at the moment of operation. Note that the sequence $$$a$$$ changes and the next operation is performed on this changed sequence.

For example, if $$$a=[3,5,4,5]$$$, then he can select $$$i = 2$$$, because $$$a_2 = 5$$$ is not divisible by $$$i+1 = 3$$$. After this operation the sequence is $$$[3,4,5]$$$.

Help YouKn0wWho determine if it is possible to erase the whole sequence using the aforementioned operation.

Input format:
The first line contains a single integer $$$t$$$ ($$$1 \le t \le 10\,000$$$)  — the number of test cases.

The first line of each test case contains a single integer $$$n$$$ ($$$1 \le n \le 10^5$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output format:
For each test case, print "YES" (without quotes) if it is possible to erase the whole sequence using the aforementioned operation, print "NO" (without quotes) otherwise. You can print each letter in any register (upper or lower).

Constraints:
Time limit: 1.0 s
Memory limit: 256.0 MB

Note:
In the first test case, YouKn0wWho can perform the following operations (the erased elements are underlined): $$$[1, \underline{2}, 3] \rightarrow [\underline{1}, 3] \rightarrow [\underline{3}] \rightarrow [\,].$$$

In the second test case, it is impossible to erase the sequence as $$$i$$$ can only be $$$1$$$, and when $$$i=1$$$, $$$a_1 = 2$$$ is divisible by $$$i + 1 = 2$$$.

Example input:
5
3
1 2 3
1
2
2
7 7
10
384836991 191890310 576823355 782177068 404011431 818008580 954291757 160449218 155374934 840594328
8
6 69 696 69696 696969 6969696 69696969 696969696

Example output:
YES
NO
YES
YES
NO

Editorial:
Notice that we can erase $$$a_i$$$ at positions from $$$1$$$ to $$$i$$$. So for each $$$i$$$, there should be at least one integer from $$$2$$$ to $$$i + 1$$$ such that $$$a_i$$$ is not divisible by that integer. If it is not satisfied for some integer $$$i$$$, then there is no solution for sure. Otherwise, it turns out that a solution always exists.
Why?
We can prove it by induction. Let's say it is possible to erase the prefix containing $$$n - 1$$$ elements. As $$$a_n$$$ can be erased at some position from $$$1$$$ to $$$n$$$(let's say $$$k$$$), then while erasing the prefix of $$$n - 1$$$ elements, when the prefix contains $$$k - 1$$$ elements, then $$$a_n$$$ is at the $$$k$$$-th position, so we can erase it at that position and erase the rest of the sequence accordingly.
So we just have to check for all integers $$$i$$$ from $$$1$$$ to $$$n$$$, if $$$a_i$$$ is not divisible by at least one integer from $$$2$$$ to $$$i + 1$$$. Notice that if $$$a_i$$$ is divisible by all integers from $$$2$$$ to $$$i + 1$$$, then it means that $$$a_i$$$ is divisible by $$$\operatorname{LCM}(2, 3, \ldots, (i+1))$$$. But when $$$i = 22$$$, $$$\operatorname{LCM}(2, 3, \ldots, 23) \gt 10^9 \gt a_i$$$. So for $$$i \ge 22$$$, there will always be an integer from $$$2$$$ to $$$i + 1$$$ which doesn't divide $$$a_i$$$. So we don't have to check for them. For $$$i \lt 22$$$, use bruteforce.
Complexity: $$$\mathcal{O}(n + 21^2)$$$.