The \text{gcdSum} of a positive integer is the $gcd$ of that integer with its sum of digits. Formally, \text{gcdSum}(x) = gcd(x, \text{ sum of digits of } x) for a positive integer $x$. $gcd(a, b)$ denotes the greatest common divisor of $a$ and $b$ — the largest integer $d$ such that both integers $a$ and $b$ are divisible by $d$.

For example: \text{gcdSum}(762) = gcd(762, 7 + 6 + 2)=gcd(762,15) = 3.

Given an integer $n$, find the smallest integer $x \ge n$ such that \text{gcdSum}(x) \gt 1.

Input

The first line of input contains one integer $t$ $(1 \le t \le 10^4)$ — the number of test cases.

Then $t$ lines follow, each containing a single integer $n$ $(1 \le n \le 10^{18})$.

All test cases in one test are different.

Output

Output $t$ lines, where the $i$-th line is a single integer containing the answer to the $i$-th test case.

Note

Let us explain the three test cases in the sample.

Test case 1: $n = 11$:

\text{gcdSum}(11) = gcd(11, 1 + 1) = gcd(11,\ 2) = 1.

\text{gcdSum}(12) = gcd(12, 1 + 2) = gcd(12,\ 3) = 3.

So the smallest number $\ge 11$ whose $gcdSum$ $\gt 1$ is $12$.

Test case 2: $n = 31$:

\text{gcdSum}(31) = gcd(31, 3 + 1) = gcd(31,\ 4) = 1.

\text{gcdSum}(32) = gcd(32, 3 + 2) = gcd(32,\ 5) = 1.

\text{gcdSum}(33) = gcd(33, 3 + 3) = gcd(33,\ 6) = 3.

So the smallest number $\ge 31$ whose $gcdSum$ $\gt 1$ is $33$.

Test case 3: $\ n = 75$:

\text{gcdSum}(75) = gcd(75, 7 + 5) = gcd(75,\ 12) = 3.

The \text{gcdSum} of $75$ is already $\gt 1$. Hence, it is the answer.

Samples

3
11
31
75

12
33
75