Difference of GCDs
You are given three integers $$$n$$$, $$$l$$$, and $$$r$$$. You need to construct an array $$$a_1,a_2,\dots,a_n$$$ ($$$l\le a_i\le r$$$) such that $$$\gcd(i,a_i)$$$ are all distinct or report there's no solution.Here $$$\gcd(x, y)$$$ denotes the greatest common divisor (GCD) of integers $$$x$$$ and $$$y$$$.
In the first test case, $$$\gcd(1,a_1),\gcd(2,a_2),\ldots,\gcd(5,a_5)$$$ are equal to $$$1$$$, $$$2$$$, $$$3$$$, $$$4$$$, $$$5$$$, respectively.
The input consists of multiple test cases. The first line contains a single integer $$$t$$$ ($$$1\le t\le 10^4$$$) — the number of test cases. The description of the test cases follows.The first line contains three integers $$$n$$$, $$$l$$$, $$$r$$$ ($$$1 \le n \le 10^5$$$, $$$1\le l\le r\le 10^9$$$).It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
For each test case, if there is no solution, print "NO" (without quotes). You can print letters in any case (upper or lower).Otherwise, print "YES" (without quotes). In the next line, print $$$n$$$ integers $$$a_1,a_2,\ldots,a_n$$$ — the array you construct.If there are multiple solutions, you may output any.
for s in [*open(0)][1:]:
    (n, l, r) = map(int, s.split())
    f = min((a := [r - r % i for i in range(1, n + 1)])) >= l
    print('NYOE S'[f::2], *a * f)
import sys
input = sys.stdin.readline
t = int(input())
for _ in range(t):
    (n, l, r) = map(int, input().split())
    res = [0] * n
    for i in range(1, n + 1):
        res[i - 1] = r - r % i
    if min(res) < l:
        print('NO')
    else:
        print('YES')
        print(*res)
for s in [*open(0)][1:]:
    (n, l, r) = map(int, s.split())
    f = min((a := [r - r % i for i in range(1, n + 1)])) >= l
    print('NYOE S'[f::2], *a * f)